Extra Credit 22

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Q16.4.15

Consider the following reaction at 298 K:

\[\ce{N2O4}(g)⇌\ce{2NO2}(g) \hspace{20px} K_P=0.142\]

What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.

Solution:

First, find standard free energy

\[\Delta G^{o}=-kTln(K_{p})\]

\[\Delta G^{o}=-8.314\frac{J}{mol\cdot k}\cdot 298K\cdot ln(0.142)\]

\[\Delta G^{o}=4.84\frac{kJ}{mol}\]

As the sign of standard free energy is positive, the reaction will favor the reactants formation.

Answer: Standard free energy change equals 4.84 KJ/mol. The reaction shifts toward the left (production of reactants increases while that of products is less). At equilibrium, Q=K and the standard free energy change is 0.

Q15.1.X

Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Table E3 for solubility products).

AgI
Ag₂SO₄
Mn(OH)₂
Sr(OH)₂•8H₂O
the mineral brucite, Mg(OH)₂

Solution:

(a)\[AgI\rightarrow Ag^{+}+I^{-}\]

	[Ag⁺]	[I^-]
Initial	0	0
Change	+x	+x
Equilibrium	x	x

\[K_{sp}=[I^{-}][Ag^{+}]\]

\[K_{sp}=x^2\]

\[1.5\times10^{^{-16}}=x^2\]

(b)\[(a)Ag_{2}SO_{4}\rightarrow 2Ag^{+}+SO_{4}^{2-}\]

	[Ag⁺]	[SO₄^2-]
Initial	0	0
Change	+2x	+x
Equilibrium	2x	x

\[K_{sp}=[Ag^{+}]^2[SO_{4}^{2-}]\]

\[(2x^2)(x)=1.2\times 10^{-5}\]

(c)\[Mn(OH_{2})\rightarrow Mn^{2+}+2OH^{-}\]

	[Mn²⁺]	[OH^-]
Initial	0	0
Change	+x	+2x
Equilibrium	x	2x

\[K_{sp}=[Mn^{2+}][OH^{-}]^{2}\]

\[(x)(2x)^2=2\times 10^{-13}\]

(d)\[Sr(OH)_{2}\rightarrow Sr^{2+}+2OH^{-}\]

	[Sr²⁺]	[OH^-]
Initial	0	0
Change	+x	+2x
Equilibrium	x	2x

\[K_{sp}=[Sr^{2+}][OH^{-}]^{2}\]

\[3.2\times 10^{-4}=x\cdot (2x)^{2}\]

(e)\[Mg(OH_{2})\rightarrow Mg^{2+}+2OH^{-}\]

	[Mg²⁺]	[OH^-]
Initial	0	0
Change	+x	+2x
Equilibrium	x	2x

\[K_{sp}=[Mg^{2+}][OH^{-}]^{2}\]

\[(x)(2x)^2=5.61\times 10^{-12}\]

Answer: (a) [Ag+]=[I-] = 9.23x10^-9 M

(b) [Ag+] = 2.88x10^-2 M [SO42-] = 1.44x10^-2 M

(d) [Sr2+] = 4.3x10^-2 M [OH-] = 8.6x10^-2 M

(e) [Mg2+] = 1.12x10^-4 M [OH-] = 2.24x10^-4 M

Q14.6.7

What is [OH⁻] in a solution of 1.25 M NH₃ and 0.78 M NH₄NO₃?

\(\ce{NH3}(aq)+\ce{H2O}(l)⇌\ce{NH4+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=1.8×10^{−5}\)

Solution:

\[K_{b}=\frac{[NH_{4}^{+}][OH^{-}]}{[NH^{3}]}\]
\[1.8\times 10^{-5}=\frac{[0.78M][OH^{-}]}{[1.25M]}\]

Answer: [OH-] = 2.88x10^-5 M

Q13.3.15

Acetic acid is a weak acid that reacts with water according to this equation:

\[\ce{CH3CO2H}(aq)+\ce{H2O}(aq)\rightleftharpoons\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)\]

Will any of the following increase the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion?

Addition of HCl
Addition of NaOH
Addition of NaCH₃CO₂

Solution:

a. Adding HCl will increase the concentration of H₃O⁺, which make the reaction procees to the left and decrease the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion

b. Adding NaOH of OH^- will decrease the concentration of H3O^{+} and let the reaction shift to the right, which will increase the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion

c. Adding NaCH₃CO₂ will increase the concentration of CH₃CO₂^- and the reaction will likely shift to the left, which will decrease the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion

Answer: The addition of a base, such as NaOH will increase the concentration of CH₃CO₂^- ions

Q5.4.43

The oxidation of the sugar glucose, C₆H₁₂O₆, is described by the following equation:

\(\ce{C6H12O6}(s)+\ce{6O2}(g)⟶\ce{6CO2}(g)+\ce{6H2O}(l)\hspace{20px}ΔH=\mathrm{−2816\:kJ}\)

The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body.

How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose?
How many Calories can be produced by the metabolism of 1.0 g of glucose?

Solution:

a.Find the amount of heat that is produced by 1.0 g of glucose

\[q=\Delta H=\frac{1.0 gC_{6}H_{12}O_{6}}{180.156 \frac{g}{mol}C_{6}H_{12}O_{6}}\times (-2816\frac{kJ}{mol})\]
b. Convert Unit from Joules to Calorie

\[-15144 Joules\times \frac{1Calorie}{4.14Joules}\]

Answer: -15.63 KJ produced -3735.79 Cal produced