Answer:
Equilibrium constant = [tex]2.23 \times 10^{83}[/tex]
Explanation:
[tex]Zr(s) + O_2(g) \rightarrow ZrO_2(s)[/tex]
[tex]E^0_{cell}[/tex] = 2.463 V
Equilibrium constant is related with [tex]E^0_{cell}[/tex] as
[tex]E_{cell}=E^0_{cell} - \frac{2.303 RT}{nF} ln k_{eq}[/tex]
In standard condition,
T = 25 °C = 25 + 273 = 298 K
F = 96500 C mol^-1
R = 8.314 [tex]J\ K^{-1}mol^{-1}[/tex]
On substituting values, the above expression becomes:
[tex]E_{cell}=E^0_{cell} - \frac{0.059}{n} log k_{eq}[/tex]
n = 2
At equilibrium, [tex]E_{cell}= 0[/tex]
[tex]0=E^0_{cell} - \frac{0.059}{2} log k_{eq}[/tex]
[tex]log k_{eq}=\frac{2 \times 2.463}{0.059}[/tex]
= 83.35
[tex]K_{eq} = antilog 83.35 = 2.23 \times 10^{83}[/tex]
