Question:
At 700 K, equilibrium constant for the reaction:

H2 (g) + I2 (g) ↔ 2HI (g)

is 54.8. If 0.5 mol L^–1of HI(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

Answer:

It is given that equilibrium constant K_cfor the reaction

H₂ (g) + I₂ (g) ↔ 2HI (g) is 54.8.

Therefore, at equilibrium, the equilibrium constant K'_cfor the reaction

2HI (g) ↔ H₂ (g) + I₂ (g) will be 1/54.8

HI = 0.5 molL^-1

Let the concentrations of hydrogen and iodine at equilibrium be x molL^-1 .

Hence, at equilibrium,

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Question:
At 700 K, equilibrium constant for the reaction:

H2 (g) + I2 (g) ↔ 2HI (g)

is 54.8. If 0.5 mol L^–1of HI(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

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Question: At 700 K, equilibrium constant for the reaction: H2 (g) + I2 (g) ↔ 2HI (g) is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

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