For the water gas reaction C(s)+H2O(g)rightharpoons CO(g)+H2(g) At 1000 K, the standard Gibbs free energy change of the reaction is -8.314 kJ/mol. Therefore, at 1000 K the equilibrium constant of the above water gas reaction is

Question

For the water gas reaction C(s)+H2O(g)rightharpoons CO(g)+H2(g) At 1000 K, the standard Gibbs free energy change of the reaction is -8.314 kJ/mol. Therefore, at 1000 K the equilibrium constant of the above water gas reaction is (A) 1 (B) 10 (C) (1/e) (D) 2.718

Tardigrade Admin · Accepted Answer

ΔrG°=-8.314×103 ∴ -ΔrG°=8.314×103=RT ln Keq or 8.314 × 103 = 8.314 ×103 ln Keq or ln Keq= 1 = ln e or Keq= e = 2.718

Q. For the water gas reaction
$C (s) + H_{2} O (g) ⇋ CO (g) + H_{2} (g)$
$A t 1000 K,$ the standard Gibbs free energy change of the reaction is $- 8.314 k J / m o l$ . Therefore, at $1000 K$ the equilibrium constant of the above water gas reaction is

$1$

$10$

$\frac{1}{e}$

$2.718$

$Δ_{r} G^{\circ} = - 8.314 \times 1 0^{3}$
$∴ - Δ_{r} G^{\circ} = 8.314 \times 1 0^{3} = RTl n K_{e q}$
or $8.314 \times 1 0^{3} = 8.314 \times 1 0^{3} l n K_{e q}$
or $l n K_{e q} = 1 = l n e$
or $K_{e q} = e = 2.718$

Q. For the water gas reaction C(s)+H2​O(g)⇋CO(g)+H2​(g) At1000K, the standard Gibbs free energy change of the reaction is −8.314kJ/mol. Therefore, at 1000K the equilibrium constant of the above water gas reaction is

1

10

e1​

2.718

Δr​G∘=−8.314×103 ∴−Δr​G∘=8.314×103=RTlnKeq​ or 8.314×103=8.314×103lnKeq​ or lnKeq​=1=lne or Keq​=e=2.718

Q. For the water gas reaction
$C (s) + H_{2} O (g) ⇋ CO (g) + H_{2} (g)$
$A t 1000 K,$ the standard Gibbs free energy change of the reaction is $- 8.314 k J / m o l$ . Therefore, at $1000 K$ the equilibrium constant of the above water gas reaction is

$1$

$10$

$\frac{1}{e}$

$2.718$

$Δ_{r} G^{\circ} = - 8.314 \times 1 0^{3}$
$∴ - Δ_{r} G^{\circ} = 8.314 \times 1 0^{3} = RTl n K_{e q}$
or $8.314 \times 1 0^{3} = 8.314 \times 1 0^{3} l n K_{e q}$
or $l n K_{e q} = 1 = l n e$
or $K_{e q} = e = 2.718$