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Characteristics of Equilibrium Constant

Calculate K...

Question

Calculate $K_{c}$ for the equilibrium process, ${C O}_{2 (g)} + H_{2 (g)} ⇌ {C O}_{(g)} + H_{2} O_{(g)}$ . At equilibrium the reaction system contains 3.52 g of ${C O}_{2}$ , 1.4 g of $C O$ , 0.08 g of $H_{2}$ and 0.72 g of $H_{2} O$ .
(Atomic masses : $C = 12, H = 1, O = 16$ )
Write up to 3 decimal places.

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Solution

${C O}_{2 (g)} + H_{2 (g)} ⇌ {C O}_{(g)} + H_{2} O_{(g)} A t {e q}^{m} : \frac{3.52}{44} \frac{0.08}{2} \frac{1.4}{28} \frac{0.72}{18} 0.08 0.04 0.05 0.04 K_{c} = K_{n} {(\frac{1}{V_{t o t a l}})}^{Δ n_{g}} = \frac{0.05 \times 0.04}{0.08 \times 0.04} {(\frac{1}{V_{t o t a l}})}^{2 - 2} = 0.625$

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Similar questions

H_{2} (g) + {C O}_{2} (g) ⇌ H_{2} O (g) + C O

2 H_{2} O (g) ⇌ 2 H_{2} (g) + O_{2} (g) K_{1} = 2.1 \times 10^{- 13}

2 {C O}_{2} (g) ⇌ 2 C O (g) + O_{2} (g) K_{2} = 1.4 \times 10^{- 12}

823

C o O (s) + H_{2} (g) ⇌ C o (s) + H_{2} O (g); K = 67

C o O (s) + C O (g) ⇌ C o (s) + {C O}_{2} (g); K = 490

2 {C O}_{2} (g) + 2 H_{2} (g) ⇌ 2 C O (g) + 2 H_{2} O (g)

C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)

H_{2} O

C O

C O

H_{2} O (g) + C O (g) ⇌ H_{2} (g) + {C O}_{2} (g)

Δ H

k J

C (g) + O_{2} (g) \to C O_{2} (g)

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

Δ H = + 131 k J

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

Δ H = - 282 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

Δ H = - 242 k J

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