For the reaction, A(g) + B(g) arrow C(g) + D(g), Δ H° and Δ S° are, respectively, -29.8 kJ mol-1 and -0.100 kJ K-1 mol-1 at 298 K. The equilibrium constant for the reaction at 298 K is :

Question

For the reaction, A(g) + B(g) arrow C(g) + D(g), Δ H° and Δ S° are, respectively, -29.8 kJ mol-1 and -0.100 kJ K-1 mol-1 at 298 K. The equilibrium constant for the reaction at 298 K is : (A) 1.0 × 10-10 (B) 1.0 × 1010 (C) 10 (D) 1

Tardigrade Admin · Accepted Answer

Δ G°=Δ H°-TΔ S° Δ G°=-29.8 KJ /mol +0.1×298 KJ mol-1 Δ G°=-2.303 RT log K If log K=0 ∴ K=1

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

$1.0 \times 1 0^{- 10}$

$1.0 \times 1 0^{10}$

10

1

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
$Δ G^{\circ} = - 29.8 K J / m o l + 0.1 \times 298 K J m o l^{- 1}$
$Δ G^{\circ} = - 2.303 RT l o g K$
If log $K = 0 ∴ K = 1$

Q. 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 298K. The equilibrium constant for the reaction at 298K is :

1.0×10−10

1.0×1010

10

1

ΔG∘=ΔH∘−TΔS∘ ΔG∘=−29.8KJ/mol+0.1×298KJmol−1 ΔG∘=−2.303RTlogK If log K=0∴K=1

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

$1.0 \times 1 0^{- 10}$

$1.0 \times 1 0^{10}$

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
$Δ G^{\circ} = - 29.8 K J / m o l + 0.1 \times 298 K J m o l^{- 1}$
$Δ G^{\circ} = - 2.303 RT l o g K$
If log $K = 0 ∴ K = 1$