Calculate the equilibrium constant for the decomposition of water 2H₂O(l) $\leftrightharpoons$ 2H₂(g) + O₂(g)
At 25^$\circ$C, given that $\Delta$ G^$\circ$_f (H₂O(l) ) = -237.2 kJ/mol.

Question

Calculate the equilibrium constant for the decomposition of water 2H₂O(l)

\leftrightharpoons

2H₂(g) + O₂(g)
At 25^$\circ$C, given that

\Delta

G^$\circ$_f (H₂O(l) ) = -237.2 kJ/mol.

Quizplus · Accepted Answer

The equilibrium constant for a reaction is given by the equation 
K = e^(-ΔG/RT), where 
ΔG = -237.2 kJ/mol (given)
R = 8.314 J/K·mol
T = 25°C = 298 K
We need to convert ΔG into Joules/mol for consistency with R, so:
ΔG = -237.2 × 1000 J/mol
Plugging in the values, we get:
K = e^(-(-237.2 × 1000)/(8.314 × 298))
K ≈ 6.3 × 10^10
Therefore, the best answer is B.

Calculate the Equilibrium Constant for the Decomposition of Water 2H2O(l) ⇋\leftrightharpoons⇋

Calculate the Equilibrium Constant for the Decomposition of Water 2H₂O(l) $\leftrightharpoons$