Question

For the equilibrium,

2H2 (g)+ Cl2 (g)<------> 2H2O (l) at 25°C ◇G° is -474.78 kJ mol^-1. Calculate log K for it . (R = 8.314JK^-1mol^-1)​

Answer 1

◇G° = -2.303 RT log K

-474.78 kJ mol^-1= -2.303×8.314 JK^-1mol^-1×298K×log K

log K =

$\frac{474.78 \times 10 ^{3}j \: mol ^{ - 1} }{2.3.3 \times 8.314 \times 298 \: j \: mol^{ - 1} }$

=+83.2

K=antilog 83.2

K=1.585×10^83

Answer 2

$\boxed{Explained\:Answer}$

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