# 18: Spontaneous Change: Entropy and Gibbs Energy

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

These are homework exercises to accompany the Textmap created for "General Chemistry: Principles and Modern Applications " by Petrucci et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

### Q18.1a

Explain how you know if each of the following change in states result in an increase or decrease in entropy.

1. solid to liquid
2. liquid to gas
3. gas to liquid

### S18.1a

1. Increase in entropy. Liquids have more entropy than solids do.
2. Increase in entropy. Gas has more entropy than liquids do.
3. Decrease in entropy. Liquids have less entropy than gases.

### Q18.1b

Determine whether each of the following is an increase or decrease in entropy.

1. fusion
2. freezing
3. combustion

1. increase
2. decrease
3. increase

### Q18.2a

Which of the following changes in states represent the smallest Delta S. Which one has the largest Delta S? How do you know?

1. freezing of water to ice
2. melting of ice to liquid water
3. sublimation of ice to gas

### S18.2a

The freezing of water is a decrease in entropy so it has the smallest Delta S because solids have less entropy than liquids. The other two options represent an increase in Entropy. The sublimation of ice is a large increase in Entropy because gas has more Entropy than solids. The sublimation of ice to gas is more of an increase than the melting of ice to water because gas has more entropy than liquid and therefore the change must have a larger Delta S.

### Q18.2c

Arrange the compounds in order of increasing entropy ($$S$$) and justify your order.

• $$CH_3OH_{(s)}$$
• $$CH_3OH_{(l)}$$,
• $$CH_3OH_{(g)}$$,
• $$CH_3CH_2OH_{(g)}$$

### S18.2c

A- CH_3 OH(s)<CH_3 OH (l)< CH_3 OH(g)< CH_3 CH_2 OH(g)

The bigger the compound, the greater the entropy as there are more degrees of freedom.

### Q18.2d

Given the reaction of diamond converting to graphite

$2C_{(s \; diamond) }→ 2 C_{(s \; (graphite)}$

Determine ∆G at 298 K and determine if this reaction is spontaneous or not. What does ∆G say about the rate of this reaction?

• $$∆H°_f (C\;(s)\; diamond)=1.9\; kJ/mol$$
• $$S° (C_{(s)}\; diamond=2.38\; J/(mol \;K)$$
• $$S° (C_{(s)}\; graphite)=5.74\; J/(mol \;K)$$

### S18.2d

$∆G=∆H-T∆S$

∆H=(2mol C(s)_(graphite)×∆H°_(f C(s)graphite) )-(2mol C(s)_diamond×∆H°_(f C(s)diamond) )

∆H=2(0)-2(1.9 kJ/mol)=-3.8kJ

∆S=(2mol C(s)_(graphite)× S_( C(s)graphite) )-(2mol C(s)_diamond× S_C(s)diamond )

∆S=2(5.74 J/(mol K))- 2(2.38 J/(mol K))=+6.72 J/mol

∆G=-3.8×10^3 J-298.15(6.72 J/mol K)=-5.51 kJ

This reaction is spontaneous. ∆G says nothing about the kinetics; while this reaction is thermodynamics favored, it occurs on a 100 million year timescale. Diamonds are kinetically meta-stable materials.

### Q18.2

Given these reactions reactions, determine whether the $$\Delta S$$ increases or decreases or stays the same at 25 °C;

1. $$H_2SO_4 (l,\; 1 atm) → H_2SO_{4 (s)} (1\; atm)$$
2. $$H_2O (l,\; 1 atm) → H_2O (g, \;1 atm)$$
3. $$CH_{4 (g)} + 2O_{2(g)} \rightleftharpoons CO_{2 (g)} + 2H_2O (l)$$

### S18.2

1. decreases because of a phase change from a liquid to a solid.
2. increases because of a phase change from a liquid to a gas.
3. decreases because there are more moles of gas in the reactants.

### Q18.3

“Spontaneous reactions are faster than non spontaneous reactions”. Access this statement and explain why it is true or false.

### S18.3

This statement is false because spontaneity does not determine speed; there are spontaneous reactions that range from very slow like melting an ice cube in cold water, to very fast like melting an ice cube in hot water.

### Q18.25a

Given the dissolving of $$CaCl_{2 (s)}$$

$CaCl_{2 (s)}→ Ca^{2+}_{(aq)} + 2Cl^-_{(aq)}$

• the $$ΔG°_f$$ of CaCl2 (s)= -748.1 kJ/mol,
• the $$ΔH$$ of CaCl2= -795.8 kJ
• the $$\Delta S^o$$
• Ca2+=-53.1 J/Kmol,
• CaCl2=104.6 J/Kmol,
• Cl-=56.5

Calculate the temperature of this reaction.

### S18.25a

$ΔG°=ΔH-TΔS$

$T=\dfrac{ΔH-ΔG°}{S}$

T=(0-[-795.8])-(-748.1))[(2(56.5)+(-53.1)]-[104.6]

T=-34.539 K

### Q18.25b

At what temperature would the following reaction occur spontaneously? (Delta H°= 1256.4 kJ; Delta S°=587 J/K)

### S18.25b

$2Cr_2O_3 \rightarrow 4Cr + 3O_2$

Delta G°= Delta H°-T DeltaS°=0

1256.4kJ=T (0.587 kJ/K) T=2140.37K

### Q18.27a

Calculate $$\Delta{G^o}$$ for

$CS_{2(l)} + 2 O_{2(g)} \rightarrow CO_{2(g)} + 2 SO_{2(g)}$

using only the following $$\Delta{G}$$ values for the reaction:

• $$C_{(s)}+O_{2 (g)}→CO_{2 (g)}$$ with $$∆G^o=-394.39\; kJ/mol$$
• $$S_{(s)}+O_{2 (g)}→SO_{2 (g)}$$ with $$∆G^o=-300.13\; kJ/mol$$
• $$C_{(s)}+2S_{(s)}→CS_{2 (l)}$$ with $$∆G^o=67.1\; kJ/mol$$

### S18.27a

A- C(s)+O_2 (g)→CO_2 (g);∆G=-394.39 kJ/mol

2[S(s)+O_2 (g)→SO_2 (g)];∆G=2[-300.13] kJ/mol

CS_2 (l)→C(s)+2S(s);∆G=-(67.1) kJ/mol

∆G= -394.39 kJ/mol + 2[-300.13] kJ/mol + -(67.1) kJ/mol = -1061.75

### Q18.27b

Use the given the standard Gibbs energy changes for these equations:

1. $$2Fe2O3 (s) \rightarrow 4Fe(s)+3O_2 (g) \;\;\;\; \Delta{G}°=-742.2 \; kJ/mol$$
2. $$Fe(OH)_3(s) \rightarrow 3Fe(s)+OH^-(g) \;\;\;\; \Delta{G}°=-696.5 \; kJ/mol$$
3. $$Fe_3O_4(s) \rightarrow 3Fe(s)+2O_2(g) \;\;\;\; \Delta{G}°=-1015\; kJ/mol$$

to identify the $$\Delta{G}°$$ for the following reaction

$6Fe_2O_3(s) \rightarrow O_2(g)+4Fe_3O_4(s)$

### S18.27b

6Fe2O3(s)-->O2(g)+4Fe3O4(s)

Fe3O4 (s) -->3Fe(s)+2O2(g) ΔG°=4(-1015) kJ/mol

2Fe2O3 (s)-->4Fe(s)+3O2 (g) -3(-742.2) kJ/mol

____________________________________________

net: 6Fe2O3(s)-->O2(g)+4Fe3O4(s) ΔG°=(-3(-742.2))+4(-1015)=3837.4 kJ/mol

### Q18.27c

Find delta G° at 298.15 K for reaction

$2CO + O_2 \rightarrow 2CO_2$

DH°= -128.3 kJ DS°=- 159.5 J K-1

### S18.27c

DG°= DH° - TDS°

Delta G°= (-128.3 kJ) – [298.15K x (-159.5 JK-1 x (1 J/1000kJ))]= -80.75 kJ

### Q18.29a

Write a the chemical equation for the complete combustion of methane, $$CH_4$$ gas and calculate ∆G° at 298.15 K.

### S18.29a

$5CH_{4(g)}+ 5O_2 (g) \rightleftharpoons 5CO_{2 (g)}+ H_2O_{(g)}$

From Tables T1 or T2

• ∆G° of $$CO_{2(g)}$$ is -393.3 kJ/mol
• ∆G° of $$H_2O_{(l)}$$ is -228.6 kJ/mol

So the standard products minus reactions approach:

∆G° = (5*[(CO)_2] + [H_2 O]) – (5*[(CH)_4) = ((5*-394.4) + (-228.6)) – (5*-50.72) ∆G° = -1947 kJ

### Q18.29

Solve for ΔG° at 298 K for the combustion of $$B_2H_{6\;(g)}$$ in $$B_2O_{3(s)}$$ and $$H_2O_{(l)}$$.

### S18.29

First the balanced reaction must be identified

$B_2H_{6(g)} + 3O_{2(g)} \rightleftharpoons B_2O_{3(s)}+ 3H_2O_{(l)}$

Using the parameters from Table T1 or T2

$ΔG°=[3(-237.1) + ( -1194)] - [86.7+3(0)] = -11,992\; kJ/mol$

The reaction is (strongly) spontaneous since $$ΔG° < 0$$ and essentially goes to completion (irreversible reaction).

### Q18.31a

Assess the feasibility of the reaction below by evaluating ($$∆S^o_{rxn}$$, $$∆H^o_{rxn}$$ and $$∆G^o_{rxn}$$) for this reaction at 25°C. If the reaction is spontaneous, is it enthalpically favored, entropically favored or both?

$(NH)_4Cl_{(s)}→ NH_{3(g)}+ HCl_{(g)}$

### S18.31a

$∆S^o = \sum(∆S^o) (products) - \sum (∆S^o) (reactants)$

(192.5 + 186.9) – (94.6) = 284.8 kJ/mol K

$∆H^o_{rxn} = \sum(∆H^o) (products) - \sum (∆H^o) (reactants)$

(-46.11 + -92.31) – (-314.4) = 175.98 kJ/mol

The Gibbs energy can be also tabulated,

$∆G^o_{rxn} = \sum(∆G^o) (products) - \sum (∆G^o) (reactants)$

but can be calculate directly from $$∆S^o_{rxn}$$ and $$∆H^o_{rxn}$$

$∆G^o_{rxn} = ∆H^o_{rxn}-T∆S^o_{rxn}$

$∆G^o_{rxn} = 175.98\; kJ/mol – 298\; K (284.8) = -84,737\; kJ/mol$

The thermal decomposition of (NH)_4Cl_{(s)} is a strongly spontaneous process that is entropically driven, but not enthalpically driven.

### Q18.31a

Given the reaction at 298 K;

$4NH_{3\;(g)} +5O_{2\;(g)} \rightarrow 4NO_{(g)}+6H_2O_{(l)}$

Determine the S

### S18.31a

ΔS°=ΔS°products-ΔS°reactants

ΔS°=(69.91+210.8)-(111.3)

ΔS°=169.41 J/molK

### Q18.31b

Assess the feasibility of the reaction:

ClNO2 (g) + NO (g) --> NO2 (g) + ClNO (g)

by calculating each of the following for the reaction at 25 °C.

1. ∆S° (standard molar entropy for ClNO2 (g) is 272.23 J/mol•K; entropy for NO (g) is 210.8 J/mol•K; entropy for NO2 (g) is 240.1 J/mol•K; entropy for ClNO (g) is 261.58 J/mol•K.)
2. ∆H° (∆Hf for ClNO2 (g) is 12.5 kJ/mol; ∆Hf for NO (g) is 90.25 kJ/mol; ∆Hf for NO2 (g) is 33.18 kJ/mol; ∆Hf for ClNO (g) is 51.71 kJ/mol.)
3. ∆G°

### S18.31b

a) ∆S° = (240.1 + 261.58) – (272.23 + 210.8) = 18.65 J / mol • K

b) ∆H° = (33.18 + 51.71) – (12.5 + 90.25) = -17.86 kJ / mol

c) ∆G° = ∆H° - T∆S° = -17860 J / mol – (298.15 K)(18.65 J / mol • K) = -23420 J / mol = -23.42 kJ / mol

∆G° < 0... therefore... reaction is spontaneous.

### Q18.37

Determine the $$K_p$$ at 298.15 K for the reaction:

$2CO_{(g)}+ O_{2(g)} \rightleftharpoons 2CO_2$

### S18.37

$∆G° = \sum (∆G°)_products - \sum ( ∆G°)_reactants$

(2*-137.2) – (2*-394.4) = 514.4 kJ/mol

$∆G^p = -RT \ln⁡ K_p$

$\ln K_p = \dfrac{∆G°}{-RT}$

- (514.4 kJ/mol)/((8.3145×10)^3*298.15 K)) = -207.5

K_p = e^(-207.5) = (7.65×10)^(-91)

### Q18.37

Given the reaction at 298 K and the thermodynamic values in Table T1

$2SO_2(g)+O_2(g) \rightleftharpoons 2SO_3(g)$

estiamte $$K_p$$.

### S18.37

$ΔG°=(2(-371.10))-(2(-300.2))$

$ΔG°=-141.8 kJ/mol$

$lnKp=-(-141.8)/((8.3145E-3)(298))$

$\ln K_p=57.23$

$K_p=e^{57.23}=7.156 \times 10^{24}$

Video Solution

### Q18.39

Determine values of ∆G° and K at 298.15 K for the following reaction:

$(2SO)_2 (g)+ O_2 (g) \rightleftharpoons (2SO)_3 (g)$

### S18.39

∆G° = ∑ (∆G°)_products - ∑( ∆G°)_reactants

(2*-371.1) – (2* - 300.2) = -141.8 kJ/mol

ln⁡K= (-∆G)/RT = ((141.8×10)^3 J/mol)/((8.3145 J/mol)(298.15 K)) = 57.2

K= e^57.2 = (6.94×10)^24

### Q18.39

Given the reaction at 298 K and the thermodynamic values in Table T1

$4NH_3(g)+5O_2(g) \rightleftharpoons 4NO(g)+6H_2O(l)$

solve for $$K_p$$.

### S18.39

ΔG°=(6(-237.6)+(4(86.55))-(4(-16.45))

=-1010.6kJ/mol

=-RTlnKp

lnKp=-(-1010.6)/((8.3145E-3)(298))

Kp=e407.87

Video Solution

### Q18.43

∆G° for the reaction below is 31.4 kJ/mol at 298.15 K. Use the thermodynamics quantity to decide in which direction the reaction is spontaneous when the concentrations of $$H_2SO_{4(aq)}$$, $$HSO^-_{4(aq)}$$, and $$H_3O^+_{(aq)}$$ are 0.2 M, 0.1 M, and 0.1 M respectively.

$H_2SO_{4 (aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + HSO^-_{4(aq)}$

### S18.43

$Q_C= \dfrac{ [H_3O^+] [HSO_4^-] }{ [H_2SO_4]} = \dfrac{ (0.1\; M)^2}{0.2\; M} = 0.05$

$∆G = 31.4 kJ + ((8.3145 \times 10)^3 \; kJ/(mol*K)) (298.15\; K) \ln⁡ 0.05= 23.97 kJ$

Because ∆G is positive, the reaction is not spontaneous going in the forward direction under these conditions.

### Q18.43

Given the ionization of ethanol in water at 298 K;

$CH_3COH_{(g)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + CH_3CO^-_{(aq)}$

when the ΔG°=28 kJ/mol and the concentration of CH3CHO(g)=0.1 M, H3O+(aq)= 0.09129 M CH3CO-(aq)=0.09129 M consider in which direction the spontaneous reaction will go.

### S18.43

Qc=0.091292(0.1-0.09129) Qc=0.949 ΔG°=28 +(8.314E-3)(298)(ln.949)=+27.87

spontaneous rxn happens in backward direction .

### Q18.47

Kp=0.12520.3352 Kp=0.139

ΔG°1000k=-RTlnK =-(8.3145J/molK)(1000K)ln0.139 =1.64 kJ/mol

Qp=(0.032)(0.022)(0.071)(0.095)

Qp=0.104

Because $$Q_p<K_p$$, the spontaneous reaction will go to the right to form more products.

### S18.45a

Kp can be calculated by using the partial pressures of each element in the system. However, since partial pressures are not given, one can use the mole fractions of products over reactants, which in this case is the same as moles of products over reactants. This is because A, B, C, and D are all divided by the same total moles.

$K_p= \dfrac{(0.6)(0.6)}{(0.4)(0.4)} = 2.25$

∆G°500K=-RTlnKp= -3.371 KJ

Qp=(.2X.017)/(.01X.3)= 1.133<2.25, which implies that the reaction will proceed to the right to reach equilibrium.

### Q18.45b

For the reaction

$NH_{3\;(aq)}+H_2O_{(l)} \rightleftharpoons NH_{4\;(aq)}^+ + OH^-_{(aq)}$

which equation is invalid.

1. $$ΔG=ΔG°+RT\ln{Q}$$
2. $$K=K_p$$
3. $$K_p=e^{-ΔG°/RT}$$

### S18.45b

C is incorrect because the correct equation is G=-RTlnKp, thus correct format would be Kp=e-GRT

### Q18.45c

For reactions involving only gases, which of the following statements is false?

1. ∆G°1/(-RT)=lnk
2. T∆S=∆H°-∆G
3. Kc=([products])/([reactants])
4. K=e^(∆G°1/RT)

### Q18.45c

D is incorrect because the exponent,∆G°1/RT, is supposed to be multiplied by negative 1.

### Q18.47

In the reaction at 1000K:

$NO_{2\;(g)}+ H_{2\;(g)} \rightleftharpoons NO_{(g)}+H_2O_{(g)}$

calculate Kp when there are 0.335 mol NO2(g), 0.335 mol H2(g), 0.125 mol NO(g), and 0.125 mol H2O(g). Solve for ΔG°at 1000K. Evaluate in what direction the spontaneous reaction will go in if there are 0.071 mol NO2(g), 0.095 molH2(g), 0.032 molNO(g), and 0.022 mol H2O(g).

### Q18.47

The imaginary reaction A + B= C + D is at equilibrium at 500 K when there is .4 mol A, .4 mol B, .6 mol C, and .6 mol D.

1. Calculate Kp at 500 K.
2. Calculate ∆G° at 500K
3. What direction would a spontaneous reaction occur if the following were the new number of moles present in the system; .01 mol A, .3 mol B, 0.2 mol C, .017 mol D?

### Q18.49a

Calculate ∆G° for each of the following equilibrium reactions at the given temperature. (Kc values not verified for accuracy. Just for practice purposes).

1. 2H2S(g)+ CH4(g)= 4H2(g) +Cs2(g) Kc=5.27X10-8 at 400K
2. SO3(g) + NO(g)→SO2(g) + NO2(g) Kc= 2.4X10-6 at 300 K
3. C3H8(g)+CH4(g)→ 2C2H6(g) Kc= 3.6X10-4 550 K

### S18.49a

$∆G°=-RT \ln K_p$

1. $$∆G°=-(8.3145)(400\; K) \ln(5.27 \times 10^{-8})= 55.7 \;KJ$$
2. $$∆G°=-(8.3145)(300\; K) \ln(2.4 \times 10^{-6})=32.2\; KJ$$
3. $$∆G°=-(8.3145)(550 \; K) \ln(3.6 \times 10^{-4})= 36.3 \;KJ$$

### Q18.49

Given the following equations solve for the

1. 2SO2(g)+2H2(g)2SO3(g) 125 C and $$K_c=28$$
2. 4NH3(g)+5O2(g)4NO(g)+6H2O(l), 332 C and $$K_c=32$$

### S18.49

1. ΔG°=-(8.3145E-3 kJ/molK)(125+273K)(ln28)=-11.02 kJ/mol
2. ΔG°=-(8.3145E-3 kJ/molK)(332+273K)(ln32)=-17.43 kJ/mol

### Q18.53A

Consider the following imaginary reaction:

$A(s) + B(g) \rightleftharpoons C(s)$

At 30 degrees, ∆G°= 40.00 KJ

Prove that the partial pressure of B(g) is very large.

What conditions could be changed to decrease the B(g) partial pressure.

### S18.53

∆G°=-RTlnKp

40 KJ=-.0083145X303XlnKp

Kp=e-15.8774

Partial pressure of B(g)=1/(e^-15.8774) which is a very large number.

To decrease the partial pressure of B(g), one must make Kp larger which can occur by decreasing the temperature.

### Q18.55

Use data from Appendix D to calculate at 298 K for the following reaction:

H2O2(g) + CH4(g) = CH3OH (g) + H2O(g)

∆S° b) ∆H° c) ∆G° d) K

### S18.55

∆S° = 239.8 J + 188.8 J – 186.3 J +188.8 J = 53.5 J

∆H° = -200.7-241.8+136.3+ 74.81 = -231.39 KJ

∆G° = - 162 – 228.6 +105.6 +50.72 = -234.28 KJ

∆G°=-RTlnKp

-234.28 KJ= -.0083145X298XlnK

K=e^944.555

### Q18.57A

What is the temperature of the following reaction if ∆G°=-53.2 kJ, ∆H°=-21.5 kJ, and ∆S°=14.7J/K?

$2NO(g)+O_2 (g) \rightarrow (2NO)_2 (g)$

### S18.57A

In this problem we are asked to calculate the temperature of the given reaction. This can be done after rearranging ∆G°=∆H°-T∆S° in order to find T.

∆G°=∆H°-T∆S°

T∆S°=∆H°-∆G°

T=(∆H°-∆G°)/(∆S°)

Find T:

T=(-21.5×(10)^3 J-(-53.2×(10)^3 J))/((14.7 J)/K)=2.6 ×(10)^3 K

### Q18.57

What must be the temperature (K) if the following reaction has ΔG°= -38.5 kJ, ΔH°= -14.6 kJ, and ΔS°= 0.152 kJ/ K?

Fe2O3 (s) + 3CO (g) → 2Fe (s) + 3CO2 (g)

### S18.57

ΔG°=ΔH°-T ΔS°

(-38.5 kJ) = (-14.6 kJ) - T(0.152 kJ/K)

-157.24 = -T

T=157.24 K

### Q18.53

For the decomposition reaction

$2MgO_{(s)} \rightleftharpoons 2Mg_{(s)}+O_{2\;(g)}$

where at 298 K

• $$H = -601.7 kJ/mol$$ and
• $$G = -569.4 kJ/mol$$

solve for $$K_p$$.

### Q18.55

Using the equation FeCO3(s) Fe(s)+CO3(aq), solve for H& S at 298K.

### Q18.54

Determine the equilibrium constant at 298 K for the following reaction:

$Mg(OH)_{2 (s)} + 2H^+_{(aq)} \rightleftharpoons Mg^{2+}_{(aq)} + 2H_2O_{(l)}$

• ΔG°: [H2O (l)]= -237.1 kJ/mol,
• ΔG°: [Mg^2+ (aq)]= -454.8 kJ/mol,
• ΔG°: [Mg(OH)2(s)]= -833.5 kJ/mol

### S18.54

ΔG°=[2[H2O(l)]+ [Mg^2+ (aq)]] - [Mg(OH)2 (s)]

ΔG°=[2[-237.1 kJ/mol]+ [-454.8 kJ/mol] - [-833.5 kJ/mol]

ΔG°=-95.5 kJ/ mol = -95.5e3 J/mol

ΔG°= -RTlnK

lnK = -ΔG°/ RT= -(-95.5e3 J/mol)/ (8.3145 J/ mol K* 298K) = 38.5

K = e^(38.5) = 5e16

### Q57A

What must the temperature be if the following reaction has ∆G°=-53.2 kJ,∆H°=-21.5 kJ,and ∆S°=14.7J/K?

$2NO_{(g)}+O_{2 (g)} \rightarrow (2NO)_{2 (g)}$

### S57A

In this problem we are asked to calculate the temperature of the given reaction. This can be done after rearranging ∆G°=∆H°-T∆S° in order to find T.

∆G°=∆H°-T∆S°

T∆S°=∆H°-∆G°

T=(∆H°-∆G°)/(∆S°)

Find T:

T=(-21.5×(10)^3 J-(-53.2×(10)^3 J))/((14.7 J)/K)=2.6 ×(10)^3 K

### Q59A

The synthesis of water occurs by the reaction H+(aq) + OH-(aq) →H2O(l) at 25°C. Using the following information, and assuming that ∆H° and ∆S° are essentially unchanged, estimate Kp at 25°C.

∆H°(H^+ )=0; ∆H°((OH)^- )=-230.0 kJ/mol; ∆H°(H_2 O)=-285.8 kJ/mol

∆S°(H^+ )=0; ∆S°((OH)^- )=-10.75 J/(mol K); ∆S°(H_2 O)=69.91 J/(mol K)

### S59A

First, determine the value of ∆G° at 25°C, from the ∆H° and ∆S° values.

∆H°=∆(H°)_f [H_2 O(l)]-∆(H°)_f [H^+ (aq)]-∆(H°)_f [(OH)^- (aq)]

=(-285.8 kJ/mol)-(0)-(-230.0 kJ/mol)=-55.8 kJ/mol H^+

∆S°=S°[H_2 O(l)]-S°[H^+ (aq)]-S°[(OH)^- (aq)]

=(69.91 J/(mol K))-(0)—10.75 J/(mol K)=80.66 J/(mol K) ( H)^+

∆G°=∆H°-T∆S°=-55.8 kJ/mol-298K×(.08066 kJ/(mol K))=-79.84 kJ/mol=-RT ln⁡(K_p )

ln⁡(K_p=(-∆G°)/RT=(79.84×(10)^3 J/mol)/(8.3145 J/(mol K)×298 K)=32.2; K_p=e^32.2=9.6×(10)^13 )

### Q59

The synthesis of water occurs by the reaction H+(aq) + OH-(aq) →H2O(l) at 25°C. Using the following information, and assuming that ∆H° and ∆S° are essentially unchanged, estimate Kp at 25°C.

∆H°(H^+ )=0; ∆H°((OH)^- )=-230.0 kJ/mol; ∆H°(H_2 O)=-285.8 kJ/mol

∆S°(H^+ )=0; ∆S°((OH)^- )=-10.75 J/(mol K); ∆S°(H_2 O)=69.91 J/(mol K)

### S59

First, determine the value of ∆G° at 25°C, from the ∆H° and ∆S° values.

∆H°=∆(H°)_f [H_2 O(l)]-∆(H°)_f [H^+ (aq)]-∆(H°)_f [(OH)^- (aq)]

=(-285.8 kJ/mol)-(0)-(-230.0 kJ/mol)=-55.8 kJ/mol H^+

∆S°=S°[H_2 O(l)]-S°[H^+ (aq)]-S°[(OH)^- (aq)]

=(69.91 J/(mol K))-(0)—10.75 J/(mol K)=80.66 J/(mol K) ( H)^+

∆G°=∆H°-T∆S°=-55.8 kJ/mol-298K×(.08066 kJ/(mol K))=-79.84 kJ/mol=-RT ln⁡(K_p )

ln⁡(K_p=(-∆G°)/RT=(79.84×(10)^3 J/mol)/(8.3145 J/(mol K)×298 K)=32.2; K_p=e^32.2=9.6×(10)^13 )

Determine the temperature at which the K_p=1.0×(10)^5 for the reaction $N_{2 (g)}+(3H)_{2 (g)} \rightarrow (2NH)_{3 (g)}\). Using the following information: ∆H°(N_2 )=0; 3∆H°(H_2 )=0; 2∆H°((NH)_3 )=-46.11 kJ/mol ∆S°(N_2 )=191.6 J/(mol K); 3∆S°(H_2 )=130.7 J/(mol K); 2∆S°((NH)_3 )=192.5 J/(mol K) ### S61 ∆H°=2∆(H°)_f [(NH)_3 (g)]-∆(H°)_f [N_2 (g)]-3∆(H°)_f [H_2 (g)] =2(-46.11 kJ/mol )-(0)-(0)=-92.22 kJ/mol ∆S°=2S°[(NH)_3 (g)]-S°[N_2 (g)]-3S°[H_2 (g)] =2(192.5 J/(mol K))-(191.6 J/(mol K))-3(130.7 J/(mol K))=-198.7 J/(mol K) ∆G°=∆H°-T∆S°=-RT ln⁡K ∆H°-T∆S°-RT ln⁡K T=(∆H°)/(∆S°-R ln⁡K ) T=(-92.22×(10)^3 J/mol)/(-198.7 J/(mol K)-8.3145 J/(mol K) ln⁡(1.0×(10)^5))=313 K ### Q61 Determine the temperature at which the $$K_p=1.0 \times 10^5$$ for the reaction: \[N_2 (g)+(3H)_2 (g) \rightleftharpoons (2NH)_3 (g)$

Using the following information:

• ∆H°(N_2 )=0; 3∆H°(H_2 )=0; 2∆H°((NH)_3 )=-46.11 kJ/mol
• ∆S°(N_2 )=191.6 J/(mol K); 3∆S°(H_2 )=130.7 J/(mol K); 2∆S°((NH)_3 )=192.5 J/(mol K)

### S61

∆H°=2∆(H°)_f [(NH)_3 (g)]-∆(H°)_f [N_2 (g)]-3∆(H°)_f [H_2 (g)]

=2(-46.11 kJ/mol )-(0)-(0)=-92.22 kJ/mol

∆S°=2S°[(NH)_3 (g)]-S°[N_2 (g)]-3S°[H_2 (g)]

=2(192.5 J/(mol K))-(191.6 J/(mol K))-3(130.7 J/(mol K))=-198.7 J/(mol K)

∆G°=∆H°-T∆S°=-RT ln⁡K

∆H°-T∆S°-RT ln⁡K

T=(∆H°)/(∆S°-R ln⁡K )

T=(-92.22×(10)^3 J/mol)/(-198.7 J/(mol K)-8.3145 J/(mol K) ln⁡(1.0×(10)^5))=313 K

### Q61

Kp = 1e6 for the following reaction:

2SO2 (g) + O2 (g)↔ 2SO3 (g)

Determine the temperature at which this Kp occurs.

T1= ?, K1= 1e6; T2= 800K, K2= 9.1e2, and ΔH°= -1.8e5 J/mol

### S18.61

ln (K2/K1) = ΔH°/R ((1/T1)/(1/t2))

ln( 9.1e2/ 1e6)= -1.8e5/ 8.3145((1/T1)/(1/800K))

1/T1= (3.2e-4)+(1.25e-3)= 1.57e-3

T1=1/1.57e-3= 6.37e2 K

### Q63A

For the reaction $C_6 H_{6(g)} \rightleftharpoons (3C)_2H_{2 (g)}$,∆H°=66.3 kJ/mol and K_p=0.214 at 298K.

1. What is Kp at 0°C?
2. At what temperature will Kp =2.50?

### S63A

a)

ln K_2/K_1 =(∆H°)/R (1/T_1 -1/T_2 )=(66.3×(10)^3 J/mol)/(8.3145 J/(mol K)) (1/298K-1/273K)=-2.45

K_2/K_1 =e^(-2.45)=0.0863 K_2=0.0863×0.214=0.018 at 273K

b)

ln K_2/K_1 =(∆H°)/R (1/T_1 -1/T_2 )=(66.3×(10)^3 J/mol)/(8.3145 J/(mol K)) (1/T_1 -1/273K)=ln 0.214/2.50=-2.458

(1/T_1 -1/298K)=(-2.458×8.3145)/(66.3×(10)^3 ) K^(-1)=-3.08×(10)^(-4) K^(-1)

1/T_1 =1/298-3.08×(10)^(-4)=3.05×(10)^(-3)-3.08×(10)^(-4)=2.74×(10)^(-3) K^(-1); T_1=365 K

### Q63

At what temperature does this reaction become spontaneous? (ΔH°= 1118.4 kJ, ΔS°= 347.2 J/K)

$Fe_3O_{4 (s)} \rightarrow 3Fe_{(s)} + 2O_{2(g)}$

### S63

ΔG°=ΔH°-T ΔS° = 0

-1118.4 kJ =-T(0.3472 kJ/ K)

T= 3221 K = 2948°C

### Q71

When referencing a table, what would you predict the normal boiling point of mercury and the vapor pressure of mercury is at 25°C?

After referencing a table with both phases of mercury as gas and liquid, we can conclude that the normal boiling point of liquid mercury (Hg) is 356.73°C = 629.88 K, and vapor pressure of mercury is about 0.002 mmHg at room temperature.

### Q74

While mercury is useful in barometers, mercury vapor can be toxic. Given that mercury has Hvap of 56.9 kJ/mol and its normal boiling point is 356.7°C, calculate the vapor pressure in mmHg at room temperature, 25°C.

### S18.74

ΔHvap/ R * ((1/T1)- (1/T2)) lnP

((56.9e3 J/ mol)/ 8.314 J/ mol K ) * ((1/298) - (1/629.7K))=

ln(12.10)=

e^(12.10)= 179,433 mmHg

### Q83

Consider the following reaction at 25°C:

H2O(g) ↔ H2O (l)

ΔGf° (kJ/ mol): H2O (l)= -237.129, H2O (g) = -228.572

ΔHf° (kJ/ mol): H2O (l)= -285.83, H2O (g) = -241.818

S°(J/ mol): H2O (l)=69.91, H2O (g) =188.825

Calculate the Keq for the given reaction.

### S18.83

ΔH°= [-285.83] - [-241.818] = -44.012kJ/ mol

ΔS°= [69.91] - [188.825] = -118.915 J/mol k

ΔG°=ΔH°-T ΔS° = -44012 J/ mol - (298K)(-118.915 J/mol K) =-8575 J/ mol= -8.575 kJ/ mol

ΔG°= -RTlnKeq

lnK = -ΔG°/ RT= -(-8557J/mol)/ (8.3145 J/ mol K* 298K) = 38.5

K = e^(3.45) = 31.6

### Q88

Calculate ΔG° (in kJ) for the following reaction in which occurs at 25.0° C.

Fe3O4 (s) → 3Fe (s) + 2O2 (g)

ΔHf (kJ/ mol): Fe3O4 (s) = -1118.4

S°(J/ mol): Fe3O4 (s) = 146.4, Fe(s)= 27.8, O2 (g) = 205.1

ΔH°= [0] - [-1118.4] = 1118.4 kJ/ mol

ΔS°= [3(27.8) +2(205.1] - [146.4] = 347.2 J/mol K

ΔG°=ΔH°-T ΔS° = 1118.4 kJ - (298K)(0.3472 kJ/mol K) = 1015 kJ/ mol

### Q104

If ΔG°> 0 for a reaction, it must also be true that: ___________________________________

### S18.104

Forward reaction is nonspontaneous at standard state and Keq<1.

### Q1

Indicate whether each of the following changes represents an increase or decrease in entropy in a system, and explain your reasoning;

1. melting ice
2. burning gasoline
3. freezing liquid bromine.

### S1

1. increases entropy, molecular bonds breaking -> greater disorder
2. increases entropy, molecules and compounds released in the burning of anything
3. decreases entropy, frozen molecules are more ordered

### Q2

Which of the following processes has the highest $$\Delta{S}$$ at 25C:

$CO_{2(s, 1 atm)} \rightarrow CO_{2(g, 10\;mmHg)}$

or

$H_2O_{(s, 1atm)} \rightarrow H_2O_{(g, 10\;mmHg)}$

### S2

Converting CO2(s, 1atm)-->CO2(g, 10mmHg) would have a much higher $$\Delta{S}$$ because it takes much more energy to convert a substance directly from a solid state to a gaseous state.

### Q3

Why is it incorrect to say that the entropy of the world decreases and the total energy of the world fluctuates?

### S3

First law of thermodynamics states that energy is neither created nor destroyed, so there is a fixed amount of energy in the universe at a given time. Also, the entropy of the universe increases for all spontaneous, naturally occurring processes, so the entropy of the universe is boundless.

### Q4

Calculate the $$\Delta{H}$$ and $$\Delta{S}$$ for the following reaction at 25C:

$4NO_{(g)} \rightarrow 2N_2O_{(g)}+O_{2(g)}$

### S3

From Table T1, the following properties are identified:

• $$NO_{(g)}$$: $$\Delta{H}$$=90.9 kJ/mol, S=210.76 kJ/mol
• $$N_2O_{(g)}$$: $$\Delta{H}$$=82.05 kJ/mol, S=219.85 kJ/mol
• $$O_{2(g)}$$: $$\Delta{H}$$=0 kJ/mol, S=205.14 kJ/mol

$$\Delta{H}$$=[Sum of $$\Delta{H}$$ (products)]-[Sum of $$\Delta{H}$$ (reactants)]

$$\Delta{S}$$=[Sum of $$\Delta{S}$$ (products)]-[Sum of $$\Delta{S}$$ (reactants)]

$$\Delta{H}$$=[(2*82.05 kJ/mol)+(0 kJ/mol)]-[(4*90.9 kJ/mol)]=-199.5 kJ/mol

$$\Delta{S}$$=[(2*219.85 kJ/mol)+(205.14 kJ/mol)]-[(4*210.76 kJ/mol)]=-198.2 kJ/mol

### Q5

Determine $$\Delta{G}$$ at 265.25 K for the reaction:

$2NO_{(g)}+O_{2(g)} \rightarrow 2NO_{2(g)}$

### S5

From Table T1, the following properties are identified:

• ($$\Delta{H}$$=-114.1 kJ,
• $$\Delta{S}$$=-146.5 J/K).

$\Delta{G}=\Delta{H}-T(\Delta{S})$

$$\Delta{G}=(-114.1 \;kJ)-(265.25 \;K)(0.1465 \;kJ/K)$$

$$\Delta{G} =-153.0\; kJ$$

### Q6

What is the Gibbs energy change $$\Delta{G}$$ for the following reaction at 25C:

$N_2+3H_2 \rightarrow 2NH_3$

### S6

From Table T1, the following properties are identified:

• $$\Delta{H}$$=-91.8 kJ and
• $$\Delta{S}$$=-198.0 J/K.

$\Delta{G}=\Delta{H}-T(\Delta{S})$

$\Delta{G}=(-91.8 \;kJ)-(298\; K)(0.1980\; kJ/K) =-32.8\; kJ$

### Q7

Which of the following is spontaneous?

1. $$\Delta{H} >0$$ and $$T(\Delta{S}>0$$
2. $$\Delta{H} <0$$ and $$T\Delta{S}>0$$
3. $$\Delta{H} >0$$, and $$T(\Delta{S}<0$$
4. $$\Delta{H} <0$$, and $$T\Delta{S}<0$$

### Q8

At what temperatures are the following reaction spontaneous:

$Br_{2(l)} \rightarrow Br_{2(g)}$

### S8

• $$\Delta{H}$$=30.91 kJ/mol,
• $$\Delta{S}$$=93.2 J/molK

For the reaction to be spontaneous:

$\Delta{G}= \Delta{H}-T\Delta{S} < 0$

(30.91 kJ/mol)-T(0.0932 J/molK <0)

T=331.6 K

### Q9

Determine $$\Delta{G}$$ for the following reaction:

$C_2H_{4(g)}+H_2O_{(l)} \rightarrow C_2H_5OH_{(l)}$

### S9

From Table T1, the following properties are identified:

$$\Delta{G}$$ C2H5OH(l)=-175 kJ/mol

$$\Delta{G}$$ C2H4(g)=68 kJ/mol

$$\Delta{G}$$ H2O(l)=-237 kJ/mol

$$\Delta{G}$$=(Sum of $$\Delta{G}$$ products)-(Sum of $$\Delta{G}$$ reactants)

$$\Delta{G}$$=-175 kJ-68 kJ-(-237 kJ)

$$\Delta{G}$$=-6 kJ

-6 kJ<0, therefore, spontaneous

### Q10

Is the following reaction spontaneous under standard conditions?

$4KClO_{3(s)} \rightarrow 3 KClO_{4(s)}+KCl_{(s)}$

### S10

From Table T1, the following properties are identified:

• $$KClO_3$$:
• $$\Delta H_f^o=-397.7\; kJ/mol$$
• $$S^o=143.1 \;J/mol K$$
• $$KClO_4$$:
• $$\Delta H_f^o=-432.8 \;kJ/mol$$
• $$S^o=151.0 \;J/mol K$$
• $$KCl$$:
• $$\Delta H_f^o=-436.7\; kJ/mol$$,
• $$S^o=82.6\; J/molK$$

Now calculate the enthalpy of the reaction (change of entropy) under standard conditions:

• heat of reaction=3(-432.8 kJ)+(-436.7 kJ)-4(397.7 kJ)=-144 kJ

Now calculate the entropy of the reaction under standard conditions:

• $$\Delta{S}$$=3S(KClO4)+S(KCl)-4S(KClO3)
• $$\Delta{S}$$=3(151.0 J/K)+(82.6 J/K)-4(143.1\; J/K)=-36.8 J/K

Now combine them together under standard conditions:

$\Delta G=\Delta H - T\Delta S$

• $$\Delta{G}=-144\;kJ -(298\; K)(-38.6\;J/K)\left(\dfrac{1\;kJ}{1000;\;J}\right)=-133\; kJ$$

since $$\Delta{G}<0$$, the reaction is spontaneous under standard conditions.

### Q11

For what temperatures will this reaction be spontaneous?

$4KClO_{3(s)} \rightarrow 3 KClO_{4(s)}+KCl_{(s)}$

### S11

The reaction will be spontaneous when $$\Delta G < 0$$. So all temperatures $$T$$ such that this inequality is satisfied:

$\Delta{G}=\Delta{H}-T(\Delta{S}) < 0$

or solving for $$T$$

$\dfrac{\Delta{H}}{\Delta{S}} < T$

$\dfrac{-144\; kJ}{(-38.6\; J)(1 kJ/1000 J)}<3731\; K$

### Q12

The following reaction occurring at 1 atm is spontaneous at ~331 K:

$Br_{2(l)} \rightarrow Br_{2(g)}$

If the temperature increased, will the reaction?

1. remain spontaneous
2. become non-spontaneous
3. approach equilibrium
4. can not say

### Q13

Use data from Appendix D to establish for the following reaction to solve for $$\Delta{G}$$:

$2N_2O_{4(g)}+O_{2(g)} \rightarrow 2N_2O_{5(g)}$

### S13

$$\Delta{G}$$=[(2*115.1 kJ/mol)]-[(2*97.54 kJ/mol)]

$$\Delta{G}$$=35.12 kJ/mol

### Q14

Using the same reaction and $$\Delta{G}$$ calculated from the above equation, solve for $$K_p$$ at 298 K.

### S14

$\Delta{G}=-RT\ln K_p$

$\ln K_p=\dfrac{\Delta{G}}{-RT}$

$\ln K_p=\dfrac{35120 J}{[(-8.314 J/mol*K)*298 K)}$

$K_p=6.98 \times 10^{-7}$

### Q15

The standard Gibbs energy change for the reaction

$NH_{3(aq)}+H_2O_{(l)} \rightarrow NH^+_{4(aq)}+OH^-_{(aq)}$

is 29.05 kJ/mol at 298 K/ use this thermodynamic quantity to decide in which direction the reaction is spontaneous when the concentrations of $$NH_{3(aq)}$$, $$NH^+_{4(aq)}$$, and $$OH^-_{(aq)}$$ are 0.10 M, 1.0*10^-3 M, and 1.0*10^-3 M, respectively.

### S15

Qc=[NH4][OH-]/[NH3]=[(1.0*10^-3M)*(1.0*10^-3 M)]/(0.10 M)

Qc=1.0*10^-5

$$\Delta{G}$$=$$\Delta{G}$$°+RTlnQc

$$\Delta{G}$$=(29050 kJ/mol)+(8.314 J/mol*K)(298 K)(ln1.0*10^-5)

$$\Delta{G}$$=525.9 kJ/mol

### Q16

For a process to occur spontaneously:

1. the entropy of the system must increase
2. the entropy of the surroundings must increase
3. both the entropy of the system and the entropy of the surroundings must increase
4. the net change in entropy of the system and surroundings considered together must be a positive quantity
5. the entropy of the universe must remain constant

### Q17

The Gibbs energy change of a reaction can be used to assess

1. how much work the system does on the surroundings
2. the net direction in which the reaction occurs to reach equilibrium
3. how much heat absorbed from surroundings
4. the proportion of the heat evolved in an exothermic reaction that can be converted to various forms of work

### Q18.29B

Calculate ΔG for a reaction at 400K if the Keq at that temperature is 200.

ΔG = -400(8.314)ln(200) = -17620 J

### Q18.37B

Determine Kp at 298 K for the reaction CO (g) + O2 (g) àCO2 (g)

(∆Gf for CO (g) = -137.2 kJ / mol; ∆Gf (CO2) = -394.4 kJ / mol)

∆G° = -RT ln Kp ln Kp = - ∆G° / RT

∆G°rxn = -394.4 + 137.2 = -257.2 kJ / mol

ln Kp = 257.2 / (8.3145)(298) Kp = 1.11

### Q18.39B

is ΔG positive or negative for this reaction?

$HCl + NaOH \rightarrow H_2O + NaCl$

### S18.39B

ΔH = - 110.27 KJ/mol

ΔG = ΔH (- sign) – T (always positive)( ΔS)

Because we go from aq to solid in this reaction we assume that ΔS is decreasing (increasing in order, so therefore with the given information it is impossible to know)

### Q18.47B

Calculate Delta G° for the reaction at 30°C

$2ZnO \rightarrow 2Zn+ O_2$

ZnO Delta Hf°= - 1034 kJ/mol S° = 145 J/molK

Zn Delta Hf° = 0 S°= 68 J/molK

O2 Delta Hf° = 0 S°= 225 J/molK

### S18.47B

Delta H°= (0) –(-1034)= 1034 kJ

Delta S°= [2(68)+1(225)]-(145)=216 J/K

Delta G°= 1034 kJ- (303K)(0.216kJ/K)=968.55 kJ

### Q18.49B

Determine $$K_c$$ or $$K_p$$ for the following reactions:

1. $$Na^+_{(aq)} + Cl^-_{(aq)} \rightleftharpoons NaCl_{(aq)}$$ with $$∆G° = -45.2\; kJ/mol\; @ \;267\; K$$
2. $$CN_{(g)} + H_{2 (g)} \rightleftharpoons HCN_{(g)}$$ with $$∆G° = -324\; kJ/mol\; @ \;312 \;K$$
3. $$SO_{2(g)} + ½O_{2 (g)} \rightarrow SO_{3 (g)}$$ with $$∆G° = -215\; kJ/mol \; @ \;417 \;K$$

### S18.49B

a) ln Kc = - ∆G° / RT = 45.2 / (8.3145)(267)

Kc = 1.02

b) ln Kp = - ∆G° / RT = 324 / (8.3145)(312)

Kp = 1.13

c) ln Kc = - ∆G° / RT = (215) / (8.3145)(417)

Kc = 1.06

### Q18.53B

Calculate the ΔS for the following fake reaction:

$A + B \rightarrow C + 2D$

S°: 160J, 205J, 213.6J, 69.9J

To solve this simply do products minus reactants:

(2(69.9) + 213.6) – (160+205) = ΔS

Δ = -11.6 J

### Q55B

Find Delta S^o, Delta H^o, and Delta G^o for the following reaction. Use Appendix D.

$2NaOH_{(aq)}+ H2SO_{4 (aq)} \rightarrow Na2SO_{4 (aq)} + H_2O_{(l)}$

### S55B

Delta S^o = 138.1 + 69.9 - (2*48.1 + 20.1) = 91.7 J/mol*K

Delta H^o = -1390 + -285.8 - (2*-470.1 +-909.3) =-1675.8+1849.5 = 173.7 kJ/mol

Delta G^o = -1268 +-237.1 - (2*-419.2+-744.5 ) = - 1505.1 + 1582.9 = 77.8 kJ/mol

### Q18.57B

Using an equation for Gibbs free energy, solve to express T as a function of delta G,H,S

T= (delta H - delta G)/delta S

### Q18.59

Find $$K_{eq}$$ at 25°C for the oxidation of iron (rusting).

$4Fe+ 3O_2 \rightarrow 2Fe_2O_3$

### S18.59B

Delta Hf° 0 0 -921kJ/mol

Delta S° 31 210 81 J/mol K

Delta G°= -RT lnK

Delta G°= Delta H°-TDeltaS°

Delta H°= [(-921)2]-[(0)4+(0)3]= -1842 kJ -1.842 x 106J

Delta S°=[(81)2]-[(31)4+(210)3]= -592 J/K

Delta G°= Delta H°- T Delta S°

=(-1.842 x 106J)-(298K)(-592 J/K)=-1665548. (spontaneous because negative)

K=e^[-DeltaG°/RT]= e^[(-1665548J)/(8.3145)(298)]

### Q18.61B

Determine the temperature, in Kelvin, at which Kp = 2.3e-5 for the following reaction, given Kp = 1.0e3 at 345 K, and ∆H°rxn = 3.2e4 J / mol.

$SO_{3 (g)}+ O_{2 (g)} \rightarrow SO_{4 (g)}$

### S18.61B

ln (2.3e-5 / 1.0e3) = (3.2e4 / 8.3145)[(1/345) – (1/T1)]

T1 = 134 K

### Q18.63B

For the reaction $N_2O_4 \rightarrow 2NO_2$ and $$K_p = 0.25$$ at 298 K

At what temperature does $$K_p = 1.00$$?

### S18.63B

First solve for Kc.

0.25 = Kc (8.314x298)^(2)

Kc = 4.0x10^-8

Plug in values for Kc and Kp

1 = 4.0x10^-8 (8.314xT)^2

T = .000024 K

### Q71B

Using the Clausius-Clapyron equation, calculate the boiling point of benzene for a pressure of 260 mmHg. The normal boiling point of Benzene is 80.1 degC and has a heat of vaporization of 30.72 kJ/mole

### S18.71B

The Clausius-Clapeyron Equation is:

Ln(P2/P1) = (∆H/R)*(1/T1-1/T2)

Substituting:

Ln(260/760) = (30720/8.314)*(1/T1 - 1/(80.1+273.15))

Solving for T1 gives T1 = 310degK -273 = 37degC

### Q18.74B

Define a in the equation S = Sinitial - R*ln(a).

a = (the effective concentration of a substance in a system) / (the effective concentration of tha substance in a standard reference state)

### Q18.83B

What temperature will the equilibrium constant for formation of $$COCl_2$$ be $$K_p= 1.6 \times 10^4$$?

$CO+ Cl_2 \rightarrow COCl_2$

### S18.83B

From Table T1 or T2

• $$\Delta H° = - 95\; kJ \; mol^{-1}$$
• $$\Delta S°=-256 \;J mol^{-1}K^{-1}$$

$\Delta G°=\Delta H°-T \Delta S°= -RT \ln K_p$

$T= \dfrac{(-9500 \;J \;mol^{-1})}{[-256 \;J\;mol^{-1}K^{-1} –(8.3145 \;J\;mol^{-1}K^{-1} \ln (1.6 \times 10^4)}$

$=\dfrac{-9500}{-336.5}= 28.23 K$

### Q18.88B

Consider the rusting of iron at 298 K:

$4 Fe_{(s)} + 3 O_{2 (g)} \\rightleftharpoons 2 Fe_2O_{3 (s)}$

1. calculate ∆G° for this reaction at 298 K
2. determine whether the reaction proceeds spontaneously in the forward or reverse direction.

Heavy rust on the links of a chain. of Wikipedia (Marlith)

### S18.88B

a) ∆S°rxn = 2(67.4) – 4(45.3) = -46.4 J / mol • K

∆G°rxn = -349000 J / mol – (-46.4)(298) = -21.1 kJ / mol

b) reaction is spontaneous in the forward direction.

### Q18.104B

if the Keq is zero, does Δ G = 0?

### S18.104B

No, Δ G would be zero if Keq =1, the equivalence point.

### Q18.3D

What is Boltzmann’s equation for entropy? Explain in detail how he derived this equation and how it shows the relationship between entropy and the number of microstates particles can occupy.

### S18.3D

S = KlnW.

S is the entropy, K is Boltzmann’s constant, and W is the number of microstates. He was able to derive this equation based on the fact that the more microstates a system has, the greater the entropy; so as the W increases, as will S.

http://chemwiki.ucdavis.edu/Physical_Chemistry/Statistical_Mechanics/Boltzmann_Average/The_Boltzmann_constant

### Q18.37D

Use data from Appendix D of Petrucci to determine $$K_p$$ at 298 K for the reaction

$2CO(g) + O_2(g) \rightleftharpoons 2CO_2(g).$

### S18.37D

$K_p = K_c(RT)^{∆n}$

$∆G° = -RT\lnK_c$

-128.6 = (.0083145)(298 K)lnKc (∆G calculated from difference of values found in Appendix D)

-128.6/2.477 = lnKc

e^-51.90 = Kc

Kc = 2.88 x 10^-23

Kp = (2.88 x 10^-23)(.0821)(298)(-1)

Kp = -7.14 x 10^-20

http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Electrochemistry_and_Thermodynamics

### Q18.49D

For the following equilibrium reactions, calculate ∆G° at the indicated temperature. [Hint: How is each equilibrium constant related to a thermodynamic equilibrium constant, K?]

1. $$HCl(g) + O_2(g) \rightleftharpoons H_2O(g) + Cl_2(g)$$ with $$K_c = 2.3 \times 10^{-12}$$ at 25°C
2. $$H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g)$$ with $$K_c = 4.5 \times 10^{-7}$$ at 25°C
3. $$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$$ with $$K_c = 1.5 \times 10^{-13}$$ at 25°C

### S18.49D

Use equation ∆G° = -RTlnK

a.) ∆G° = -(.008314)(298 K)ln(2.3 x 10^-12)

∆G° = 66.45 kJ/mol

b.) ∆G° = -(.008314)(298 k)ln(4.5 x 10^-7)

∆G° = 36.207 kJ/mol

c.) ∆G° = (.008314)(298 K)ln(1.5 x 10^-13)

∆G° = 73.15 kJ/mol

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Free_Energy/Gibb's_Free_Energy/Gibbs_Free_Energy

### Q18.61D

Use the equations ∆G = ∆H –T∆S and ∆G° = -RTlnK, and Appendix D to estimate the temperature at which Kp = 2.6 x 10^3 for the reaction 2SO2(g) + O2(g)↔2SO3(g).

### S18.61D

Assume ∆G° = -70.9 (from appendix D, difference of both ∆G° values)

-70.9 = -(.0083145)T(ln(2.6 x 10^3))

T = approximately -10844.4 K

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Free_Energy/Gibb's_Free_Energy/Gibbs_Free_Energy

### Q18.88D

The standard molar entropy of solid hydrazine at its melting point of 1.53°C is 67.15 J/mol*K. The enthalpy of fusion is 12.66 kJ/mol. For N2H4(l) in the interval from 1.53°C to 298.15K, the molar heat capacity at constant pressure is given by the expression Cp = 97.8 + 0.0586(T – 280). Determine the standard molar entropy of $$N_2H_{4(l)}$$ at 500 K. [Hint: The heat absorbed to produce an infinitesimal change in temperature of a substance is ∆qrev= (Cp)(∆T)].

### S18.88D

Cp = 97.8 + .0586(500-280)

Cp = 110.7

∆qrev = (110.7)(23.47 K) = 2598.13

∆S = qrev/T

∆S = 2598.13/500 K = 5.196 J/K

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Entropy

### Q18.1E

Is the entropy increasing or decreasing?

1. Tear open and start using a cold pack.
2. Boiling water.
3. Freezing water.

### S18.1E

(a) Increasing (b) Increasing (c) Decreasing

### Q18.2E

Put them in order of increasing △S at 25℃.

1. CO2 (g, 1atm) ↔ CO (s, 1atm)
2. CO2 (s, 1atm) ↔ CO2 (l, 1atm)
3. H2O (s, 1atm) ↔ H2O (g, 1atm)

### Q18.27E

Calculate △G° for the reaction

$COS_{(s)} + H_2O (g) \rightleftharpoons CO_{2\; (g)} + H_2S_{(g)}$

given the known thermodynamic properties of the following reactions

1. $$CO_{(g)} + H_{2\;(g)} \rightleftharpoons COS_{(g)} + H_{2 (g)}$$ △G° = 14.7 kJ
2. $$CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2\; (g)} + H_{2 \; (g)}$$ △G° = -28.6 kJ

△G° = -27.2 kJ.

### Q18.29E

What’s △G° For the reaction

$2SO_{2\; (g)} + O_{2\;(g)} \rightleftharpoons 2SO_{3\; (g)}$

△G°f of $$SO_2$$ is -300kJ/mol, and its -371 kJ/mol of $$SO_3$$.

△G° = -142kJ.

### Q18.31E

Find △H° if the reaction Fe3O4(s) ↔ 3Fe(s) + 2O2(g) happens spontaneously. The temperature is 2948℃,△S° = 347.2 J/K.

△H° = 1118.3 kJ.

### Q18.37E

If the Keq for the reaction CaCO3(s) ↔ CaO(s) + CO2(g) is 8.1x10-5, find △G when T=25℃. Given: [CaCO3] = 0.003M, [CaO] = 0.0004M, and [CO2]=0.0005M.

△G = -482.5 J.

### Q18.39E

What’s the △S° of the reaction 2NO(g) + O2(g) ↔ 2NO2(g)? If the △S° is 210.8 J/K , 205.1 J/K, and 240.1 J/K respectively.

### S18.39E

△S° = -146.6 J/K.

### Q18.43E

For the reaction, N2O(g) + ½ O2(g) ↔ 2NO2(g), the partial pressure are 0.1atm, 0.5atm, and 0.8atm respectively. Is the reaction spontaneous?

### S18.43E

It’s nonspontaneous.

### Q18.45E

For reaction 3Fe2O3(s) +H2(s) ↔ 2Fe3O4(s) + H2O(g), which of the following is correct and why?

1. △G = △G° + RT lnQ.
2. K=-kp.
3. Kp = eG°/RT

A

### Q18.47E

The gas concentrations of the reaction

$2A_{(g)} + 3B_{(g)} \rightleftharpoons 2C_{(g)} + 4D_{（g）}$

are [A] = 1.8x10-2M, [B] = 3.4x10-3M, [C] = 5.6x10-1M, [D]=2.8x10-4M. Assume T=25 ℃, and $$K_{eq} = 4.42 \times 10^{-9}$$. Find Q, △G°, and △G.

### S18.47E

1.51x10-4J, 4.77x104J, 25.9kJ

### Q18.49E

At 298K, Mg(OH)2 (s) + 2H+(aq) ↔ Mg2+(aq) + 2H2O(l) has △G° = -95.5kJ/mol. If the Q is 2.389x10-4, what’s △G?

### S18.49E

△G = 1.97x106 kJ.53.

### Q18.53E

Find the $$K$$, $$K_c$$, and $$K_{eq}$$ of the reaction

$Si(s) + 2Cl_2(g)\rightleftharpoons SiCl4(g)$

### S18.53E

K= [SiCl]/[Cl2-]2=Kc=Ksp.

### Q18.55E

If T has a positive value, △H has a negative value, △S has a negative value, when will △G be positive or negative?

### S18.55E

△G will be positive/negative when △H is smaller/greater than T△S

### Q18.57E

If T has a positive value, △H has a negative value, △S has a positive value, when will △G be positive or negative?

### S18.57E

△G will be positive/negative when △H is greater/smaller than T△S.

### Q18.59E

What’s the temperature of the reaction

$H_2(g) +I(g) \rightleftharpoons 2HI（g）$

if $$\Delta{H}° = -30kJ$$, $$\Delta{S}° = 30 J/K$$, and $$K_{eq} = 2.5 \times 10^{-3}$$?

Video Solution

T= 855.8K

### Q18.61E

Given K2=2, K1=3. Temperature increases from 100℃ to 120℃. What’s △H°?

△H° = 358.8J.

### Q18.63E

For the reaction

$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$

calculate △G at 25℃ if it’s known that △Gf° (SO3) = -150kJ, △Gf° (SO2) = -125kJ, Q= 2.0x10-3

△G= 159.6 kJ.

### Q18.71E

For problem 61, if the value of Q is not given, instead, it says that SO2 gas is at 2atm, O2 is at 0.5atm, and SO3 gas at 4 atm. Calculate the new △G.

△G = 176.7 kJ.

### Q18.74E

For the reaction, Br2(l) ↔ Br2(g), what’s the normal boiling point of it says that △H° = 31.0kJ/mol, △S° = 93.07 J/Kmol?

△H° = 333K.

### Q18.83E

Calculate △G° for the oxidation of iron:

$4Fe(s) +3O_2(g) \rightleftharpoons 2Fe_2O_3(s)$

given:

• △Hf° of $$2Fe_2O_3(s)$$ is -826 kJ/mol,
• Sm° of $$Fe­_{(s)}$$ is 27 J/mol K and 205 J/mol K for $$O_2$$ and 90 J/mol K for $$Fe_2O_{3(s)}$$ .

△G°= 10261J.

### Q18.88E

What would happen if increasing the temperature of the reaction

$H_{2\; (g)} + I_{2\; (g)} \rightleftharpoons 2HI_{(s)}$

? Given that $$K_p = 620$$ at 298 K and $$△H° = -9.5 \; kJ/mol$$.

### S18.88E

products are favored.

### Q18.104E

For the reaction

$2SO_{2\; (g)} + O_{2\; (g)} \rightleftharpoons 2SO_{3\; (g)}$

if it’s known that $$K_p = 500$$ at 25 ℃ and $$△H° = -7.23\; kJ/mol$$, what would happen if we decrease the temperature?

### S18.104E

the reactants are favored.

## Misc

### Q1

Using the following values for entropy determine if a reaction would be spontaneous.

I) ΔSsys = 30 J/K , ΔSsurr = 50 J/K

II) ΔSsys = 60 J/K, ΔSsurr = -85 J/K

A reaction will be spontaneous if the total entropy change is positive. It will not be spontaneous if the total entropy change is negative.

ΔStotal = ΔSsys + ΔSsurr

where

ΔStotal is the total entropy change

ΔSsys is the entropy change of the system

ΔSsurr is the entropy change of the surroundings

I.) ΔStotal = ΔSsys + ΔSsurr

ΔStotal = 30 J/K + 50 J/K

ΔStotal = 80 J/K

The total entropy change is positive, therefore the reaction will be spontaneous.

II.) ΔStotal = ΔSsys + ΔSsurr

ΔStotal = 60 J/K + -85 J/K

ΔStotal = -25 J/K

The total entropy change is negative, therefore the reaction will not be spontaneous.

### Q2

Under what conditions are the following reactions spontaneous, and why?

1. 3N2(g) ⇄ 2N3(g)
2. H2O(l) H2O(g) where ∆H is negative
3. 2NH4NO2(s) 2N2(g) + 4H2O(g) + O2(g) where ∆H is negative

### S2

1. Non spontaneous at all temperatures because entropy is decreased
2. Spontaneous when temperature is high, above 100˚C at STP
3. This reaction is spontaneous at all temperatures because of negative ∆H and increasing entropy of the system

### Q25

Calculate the change in Gibbs free energy.

HCl+NaOCH3 → NaCl+HOCH3

pKa HCl=-7pKa HOCH3=16 T=298K

Keq=[NaCl][HOCH3][H+]/[HCl][NaOCH3][H+]

Keq=KaHCl/KaHOCH3

Keq=10^7/10^-16= 10^23

∆G=-RTlnK

∆G=-8.314x298xln10^23

∆G=-13.1kJ ← favored towards products

### Q35

In the synthesis of gaseous methanol from carbon monoxide gas and hydrogen gas, the following equilibrium concentrations were determined at 444 K: [CO(g)]=0.0977M, [H2O(g)]=0.0799M, and [CH3OH(g)]=0.00799M. Calculate the equilibrium constant and Gibbs energy for this reaction.

### S35

CO(g)+2H2(g) → CH3OH(g)

Keq=[CH3OH]/([CO][H2]^2)

Keq=[0.00799]/([0.0977][0.0799]^2)

Keq=12.8

ΔGº=-RTln(Keq)

ΔGº=-(8.314J/Kmol)(444K)ln(12.8)(kJ/1000J)

ΔGº=-9.4kJ/mol

### Q37

For the following reaction, what would ΔGº be at 298K ?

Fe3O4(s) → 3 Fe (s) + 2O2(g)

ΔHfº (kj/mol)-1118.4

ΔSº (J/molK) 146.427.8205.1

ΔHº 0 - (-1118.4) = +1118.4 kJ

ΔSº (3(27.8) + 2(205.1)) - 146.4 = 347.2 J/K

ΔGº = ΔHº-TΔSº = 1118.4kJ-(298.15K)(.3472kJ/K)= 1015kJ

### Q41

If delta H = 158 kJ and delta S = 411 J/k. At what temperature will this reaction be spontaneous?

### S41

deltaG= deltaH-TdeltaS

0> 158000J – T(411 J/K)

T(411 J/K)/ 411 J/K > 158000 J/ 411 J/K

T > 384 K

### Q55

Calculate ∆G of this reaction

∆H=-537.22kJ∆S=13.74J/KT=25ºC

H2(g)+F2(g) → 2HF

∆G=∆H-T∆S

∆G=(-537.22kJ)-(298K)(13.74J/K)

∆G=(-537.22kJ)-(298K)(.01374kJ/K)

∆G=-541.31kJ ← SPONTANEOUS

### Q57

What must be the temperature if the following reaction has ΔGº=-52.9 kJ, ΔHº=-34.7kJ, and ΔSº=12.4J/K?

Fe2O3(s)+3CO(g) → 2Fe(s)+3CO2(g)

ΔGº=ΔHº-TΔSº

(-52.9 kJ)(1000J/kJ)=(-34.7 kJ)(1000J/kJ)-T(12.4J/K)

T=1470K

### Q67

Label which reaction is spontaneous or nonspontaneous, and compute the overall reaction, given that it is spontaneous.

Cu2O (s) → 2 Cu(s) + ½ O2 (g) ΔGº = 125 kJ

C(s) + ½ O2(g) → CO(g) ΔGº = -175

Reaction 1: nonspontaneous; reaction 2: spontaneous

Net reaction:

Cu2O (s) + C (s) → 2 Cu (s) + CO(g)

ΔGº= 125+ (-175) = -50 kJ

Net reaction is spontaneous.

18: Spontaneous Change: Entropy and Gibbs Energy is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.