For the reaction, 2NH3(g)+CO2(g) leftharpoons NH2CONH2(aq)+H2O(l) Standard Gibbs energy change, ΔrG° at the given temperature is -13.6 KJ mol-1 The value of equilibrium constant for the above reaction at 298 K is

Question

For the reaction, 2NH3(g)+CO2(g) leftharpoons NH2CONH2(aq)+H2O(l) Standard Gibbs energy change, ΔrG° at the given temperature is -13.6 KJ mol-1 The value of equilibrium constant for the above reaction at 298 K is (A) 8.5 × 102 (B) 2.4 × 102 (C) 3.5 × 10-2 (D) 4.3 × 10-7

Tardigrade Admin · Accepted Answer

We know , log K =(-Δ G°/2.303RT) =-((-13.6 × 103 J mol-1)/2.303(8.314 JK-1 mol-1)(298 K))=2.38 Hence, K = antilog 2.38=2.4 × 102

Q. For the reaction,
$2 N H_{3 (g)} + C O_{2 (g)} ⇌ N H_{2} CON H_{2 (a q)} + H_{2} O_{(l)}$
Standard Gibbs energy change, $Δ_{r} G^{\circ}$ at the given temperature is $- 13.6 K J m o l^{- 1}$ The value of equilibrium constant for the above reaction at $298 K$ is

$8.5 \times 1 0^{2}$

$2.4 \times 1 0^{2}$

$3.5 \times 1 0^{- 2}$

$4.3 \times 1 0^{- 7}$

We know , $l o g K = \frac{- Δ G ^{\circ}}{2.303 RT}$
$= - \frac{( - 13.6 \times 1 0 ^{3} J m o l ^{- 1} )}{2.303 ( 8.314 J K ^{- 1} m o l ^{- 1} ) ( 298 K )} = 2.38$
Hence, $K$ = antilog $2.38 = 2.4 \times 1 0^{2}$

Q. For the reaction, 2NH3(g)​+CO2(g)​⇌NH2​CONH2(aq)​+H2​O(l)​ Standard Gibbs energy change, Δr​G∘ at the given temperature is −13.6KJmol−1 The value of equilibrium constant for the above reaction at 298K is

8.5×102

2.4×102

3.5×10−2

4.3×10−7