Given the following equilibrium equations and their corresponding equilibrium constants: 2CO2 (g)+H2O(g)⇌2O2 (g)+CH2CO(g) Kc=6.1x108 CH4(g)+2O2(g)⇌CO2 (g)+2H2O(g)...

Question

Question

Given the following equilibrium equations and their corresponding equilibrium constants:

2CO2 (g)+H2O(g)⇌2O2 (g)+CH2CO(g) Kc=6.1x108 CH4(g)+2O2(g)⇌CO2 (g)+2H2O(g) Kc=1.2x1014 Find Kc for the reaction: CH4(g) + CO2(g) ⇌ CH2CO (g) + H2O (g)

Answer 1

Answer #1

For given problem first we need to add up the two reaction.

2 CO₂ (g) + H₂O (g) $\rightleftharpoons$ 2 O₂ (g) + CH₂CO (g) K_c = 6.1 * 10⁸

+ CH₄ (g) + 2 O₂ (g) $\rightleftharpoons$ CO₂ (g) + 2 H₂O (g) K_c = 1.2 * 10¹⁴

Cancelling 2 O₂ (g) and subtracting H₂O (g) and CO₂ (g) from both side, we get

CH₄ (g) + CO₂ (g) $\rightleftharpoons$ CH₂CO (g) + H₂O (g).

To find the K_c for combined reaction we need to multiply both the value.

Therefore K_c = 6.1 * 10⁸ * 1.2 * 10¹⁴ = 7.32 * 10²².

After rounding off final answer will be 7.3 * 10²².

Therefore K_c for reaction CH₄ (g) + CO₂ (g) $\rightleftharpoons$ CH₂CO (g) + H₂O (g) is 7.3 * 10²².

If you find any mistake please mention in the comment box.

Thanks.

Answer 2

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