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Decomposition Reaction

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Question

Find out the value of equilibrium constant for the following reaction at $298 K$ .
$2 N H_{3} (g) + C O_{2} (g) ⇌ N H_{3} C O N H_{2} (a q) + H_{2} O (l)$ Standard Gibbs energy change, $Δ_{r} G^{⊖}$ at the given temperature is $- 13.6 k J m o l^{- 1}$ .

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Solution

Equilibrium constant

We know , $\begin{matrix} log K & = \frac{- Δ_{r} G^{⊖}}{2.303 R T} = \frac{- (- 13.6 \times 10^{3} J m o l^{- 1})}{2.303 (8.314 J K^{- 1} m o l^{- 1}) (298 K)} H e n c e K & = antilog 2.38 = 2.4 \times 10^{2} \end{matrix}$

Final Answer : $2.4 \times 10^{2}$

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Similar questions

2 N H_{3 (g)} + C O_{2 (g)} ⇌ N H_{2} C O N H_{2 (a q)} + H_{2} O_{(l)}

Δ_{r} G^{\circ}

- 13.6 K J m o l^{- 1}

^{- 1}

^{- 1}

^{- 1}

K_{c}

Δ G^{o} in k J m o l^{- 1}

K

Δ_{f} H^{o}

C O_{2} (g), C O (g)

H_{2} O (l)

- 393.5, - 110.5

- 241.8 {kJ mol}^{- 1}

{kJ mol}^{- 1}

298 K

C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (l)

R = 8.3 J/K/mol

I_{2} + H_{2} S \to 2 H I + S

△ G_{f}^{o} (H I) = 1.8 k J m o l^{- 1}, △ G_{f}^{o} (H_{2} S) = 33.8 k J m o l^{- 1}

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