For the reaction A(g)+B(g)→C(g)+D(g) ΔH∘ and ΔS∘ are, respectively, −29.8kJmol−1 and −0.100kJK−1mol−1 at 298 K. The equilibrium constant for the reaction at 298 K is:

For the reaction $A (g) + B (g) \to C (g) + D (g) Δ H^{\circ} and Δ S^{\circ}$ are, respectively,
$- 29.8 kJmol - 1$ and $- 0.100 {kJK}^{- 1} {mol}^{- 1}$ at 298 K. The equilibrium constant for the reaction at 298 K is:

A
$1.0 \times 10^{- 10}$
B
10
C
1
D
$1.0 \times 10^{10}$

Solution:

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = (- 29.8 kJ {mol}^{- 1}) - (298 K) (- 0.100 {kJK}^{- 1} {mol}^{- 1}) = 0$

From the expression, $Δ G^{\circ} = - R T \ln K_{p}^{\circ}$ , we get $K_{p}^{\circ} = 1$ .

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