Extra Credit 22
- Page ID
- 96906
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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.
\[\Delta G^{o}=-8.314\frac{J}{mol\cdot k}\cdot 298K\cdot ln(0.142)\]
\[\Delta G^{o}=4.84\frac{kJ}{mol}\]
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+] |
[SO42-] |
|
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
(c) [Mn2+] = 3.68x10^-5 M [OH-] = 7.36x10^-5 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