Question 16.7: Graded When zinc is dissolved in a strong acid, zinc ions an...
GRADED
When zinc is dissolved in a strong acid, zinc ions and hydrogen gas are produced.
Zn(s) + 2H^{+}(aq) → Zn^{2+}(aq) + H_{2}(g)
At 25°C, calculate
ⓐ ΔG°.
ⓑ ΔG when P_{H_{2}} = 750 mm Hg, [Zn^{2+}] = 0.10 M, [H^{+}] = 1.0 × 10^{-4} M.
ⓒ the pH when ΔG = -100.0 kJ, P_{H_{2}} = 0.933 atm, [Zn^{2+}] = 0.220 M and the mass of Zn is 155 g.
ⓐ
| ANALYSIS | |
| equation for the reaction (Zn(s) + 2H^{+}(aq) → Zn^{2+}(aq) + H_{2}(g))
T(25°C) |
Information given: |
| ΔG_{f}° values at 25°C (Appendix 1) | Information implied |
| ΔG° | Asked for |
STRATEGY
1. Find ΔG° by substituting ΔG_{f}° values into Equation 16.3.
ΔG°_{reaction} = ΣΔG°_{f_{ products}} – ΣΔG°_{f_{ reactants}} (16.3)
2. Recall that ΔG_{f}° for elements in their native state at 25°C and H^{+}(aq) is zero.
ⓑ
| ANALYSIS | |
| P_{H_{2}} (750 mm Hg); [Zn^{2+}](0.10 M); [H^{+}] (1.0 × 10^{-4} M); T(25°C) from part (a): ΔG°(-147.1 kJ) | Information given: |
| R value in energy units | Information implied |
| ΔG | Asked for |
STRATEGY
ⓒ
| ANALYSIS | |
| P_{H_{2}}(0.933 atm); [Zn^{2+}](0.220 M); mass Zn(155 g); T(25°C) ΔG(-100.0 kJ) From part (a): ΔG°(-147.1 kJ) |
Information given: |
| R value in energy units | Information implied |
| pH | Asked for |
STRATEGY
1. Find Q by substituting into Equation 16.4.
ΔG = ΔG° + RT ln Q (16.4)
2. Write the equation for Q.
3. Solve for [H^{+}] by substituting into the Q expression. Change [H^{+}] into pH.
ⓐ
| ΔG° = ΔG°_{f}Zn^{2+}(aq) + ΔG°_{f} H_{2}(g) + [ΔG_{f} Zn(s) + 2(ΔG°_{f} H^{+}(aq))] = -147.1 kJ + 0 – [0 + 2(0)] = -147.1 kJ |
ΔG° |
ⓑ
| Q = \frac{[Zn^{2+}]P_{H_{2}}}{[H^{+}]²} = \frac{(0.10) (750/760)}{(1.0 × 10^{-4})²} = 9.9 × 10^{6} | 1. Q |
| ΔG = ΔG° + RT ln Q = -147.1 kJ + (0.00831 kJ/K)(298 K) ln (9.9 × 10^{6}) = -147.1 kJ + 39.9 kJ = -107.2 kJ |
2. ΔG |
ⓒ
| –100.0 kJ = –147.1 kJ + (0.00831 kJ/K)(298 K)ln Q
ln Q = \frac{47.1}{(0.00831) (298)} = 19.0 → Q = e^{19.0} = 1.82 × 10^{8} |
1. Q |
| Q = \frac{[Zn^{2+}](P_{H_{2}})}{[H^{+}]²} = \frac{(0.220) (0.933)}{[H^{+}]²} = 1.82 × 10^{8} | 2. Q expression |
| [H^{+}] = \left(\frac{(0.220)(0.933)}{1.82 × 10^{8}} \right)^{1/2} = 3.35 × 10^{-5} M; pH = –log_{10} 3.35 × 10^{-5} = 4.47 | 3. [H^{+}]; pH |