Question

Following reaction takes place at $(\frac{600}{2.303\times0.8314})$ K temperature.

$H_2(g) + 2Ag^+ (aq) \rightleftharpoons 2Ag(s) + 2H^+(aq)$

$P_{H_2} = 1.0bar, [Ag^+] = 10^{-6}M, [H^+] = 10^{-3}M, \Delta_f G^{\circ}(Ag^+, aq) = 75 \text{ kJ mol}^{-1}$

Calculate z.

where $z = \frac{|\Delta_r G(\text{in kJ})|}{19}$

Answer 1

z=6

Answer 2

1. Standard Reaction Free Energy:
   For the reaction

   $$H_{2}(g)+2Ag^+(aq)\longrightarrow 2Ag(s)+2H^+(aq)$$

   Using $\Delta_fG^\circ(H_2)=\Delta_f G^\circ(Ag)=\Delta_f G^\circ(H^+)=0$ and given $\Delta_fG^\circ(Ag^+) = 75\,\text{kJ/mol}$,

   $$\Delta_rG^\circ = [2\times 0 + 2\times 0] - [0 +2\times (75)] = -150\,\text{kJ}$$

2. Reaction Quotient $Q$:
   With $P_{H_2}=1.0\,\text{bar}$, $[Ag^+]=10^{-6}\,M$ and $[H^+]=10^{-3}\,M$, note that solids are omitted. Thus,

   $$Q=\frac{[H^+]^2}{[Ag^+]^2\, P_{H_2}} = \frac{(10^{-3})^2}{(10^{-6})^2\times 1} = \frac{10^{-6}}{10^{-12}}=10^{6}$$

3. Temperature:
   Given

   $$T=\frac{600}{2.303\times 0.8314}\,K \approx \frac{600}{1.9124}\,K \approx 314\,K.$$

4. Non‐Standard Free Energy Change:
   Use

   $$\Delta_rG = \Delta_rG^\circ + RT\ln Q$$

   Convert $R$ to $\text{kJ mol}^{-1}K^{-1}$: $R=0.008314\,\text{kJ mol}^{-1}K^{-1}$. Then,

   $$RT = 0.008314\times 314 \approx 2.611\,\text{kJ/mol},$$ $$\ln Q = \ln(10^6)=6\ln10\approx 6\times2.303=13.818,$$ $$RT \ln Q \approx 2.611\times 13.818 \approx 36.07\,\text{kJ}.$$

   Thus,

   $$\Delta_rG = -150\,\text{kJ}+36.07\,\text{kJ}=-113.93\,\text{kJ}.$$

   Taking the absolute value: $|\Delta_rG|\approx 113.93\,\text{kJ}$.

5. Calculation of $z$:

   $$z=\frac{|\Delta_rG|}{19}=\frac{113.93}{19}\approx6.$$