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At 700 K, Equilibrium Constant for the Reaction 54.8. If 0.5 Moll–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 700 K - CBSE (Science) Class 11 - Chemistry

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Question

At 700 K, equilibrium constant for the reaction

`H_(2(g)) + I_(2(g)) ↔ 2HI_(g)` is 54.8. If 0.5 molL–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 700 K?

 

Solution

It is given that equilibrium constant `K_C` for the reaction

`H_(2(g)) + I_(2(g)) ↔ 2HI_((g))` is 54.8.

Therefore, at equilibrium, the equilibrium constant `K'_C` for the reaction

2HI_(g) ↔ H_(2(g)) + I_(2(g)) will be `1/54.8`

[HI] = 0.5 `molL^(-1)`

Let the concentrations of hydrogen and iodine at equilibrium be x molL–1

`[H_2] = [I_2] = xmolL^(-1)`

Therefore, `([H_2][I_2])/[HI]^2 = K'_C`

`=> (x xx x)/(0.5)^2 = 1/54.8`

`x^2 = 0.25/54.8`

`=> x = 0.06754`

x = 0.068 `molL^(-1)` (approximatley)

Hence, at equilibrium, `[H_2] = [I_2] = 0.068 mol L^(-1)`

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Solution At 700 K, Equilibrium Constant for the Reaction 54.8. If 0.5 Moll–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 700 K Concept: Law of Chemical Equilibrium and Equilibrium Constant.
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