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NCERT Exemplar · Q28

Q.The value of Kc for the reaction 2HI

(g) ⇌ H2
(g) + I2
(g) is 1 × 10^-4
At a given time, the composition of reaction mixture is
[HI] = 2 × 10^-5 mol, [H2] = 1 × 10^-5 mol and [I2] = 1 × 10^-5 mol
In which direction will the reaction proceed?
Jharkhand JacShort· 2mImportance★★★★★est
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Compare the reaction quotient QcQ_c with the equilibrium constant KcK_c. Here Qc=[H2][I2][HI]2=(1×10−5)(1×10−5)(2×10−5)2=0.25≫Kc=1×10−4Q_c = \frac{[H_2][I_2]}{[HI]^2} = \frac{(1\times10^{-5})(1\times10^{-5})}{(2\times10^{-5})^2} = 0.25 \gg K_c = 1 \times 10^{-4}, so the reaction proceeds in the reverse direction (toward HI) to reach equilibrium.

The equilibrium constant KcK_c tells us the ratio of product to reactant concentrations when a reaction is at equilibrium. But what happens when the system is not at equilibrium? We need to compare where the system is now with where it wants to be.

The tool for this comparison is the reaction quotient QcQ_c, which has exactly the same mathematical form as KcK_c but uses the current concentrations instead of equilibrium concentrations. By comparing QcQ_c with KcK_c, we can predict which way the reaction will shift:

  • If Qc<KcQ_c < K_c: too few products relative to equilibrium → reaction goes forward
  • If Qc>KcQ_c > K_c: too many products relative to equilibrium → reaction goes backward
  • If Qc=KcQ_c = K_c: the system is already at equilibrium → no net change

Let me work through this problem step by step.

Solution

  1. Write the expression for the reaction quotient.

    For the reaction 2HI (g)⇌H2(g)+I2(g)2\text{HI (g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)}, the reaction quotient is:

Qc=[H2][I2][HI]2Q_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}

  1. Substitute the given concentrations.

    We have:

    • [HI]=2×10−5[\text{HI}] = 2 \times 10^{-5} mol/L
    • [H2]=1×10−5[\text{H}_2] = 1 \times 10^{-5} mol/L
    • [I2]=1×10−5[\text{I}_2] = 1 \times 10^{-5} mol/L

Qc=(1×10−5)(1×10−5)(2×10−5)2Q_c = \frac{(1 \times 10^{-5})(1 \times 10^{-5})}{(2 \times 10^{-5})^2}

  1. Calculate QcQ_c.

Qc=1×10−104×10−10=14=0.25=2.5×10−5Q_c = \frac{1 \times 10^{-10}}{4 \times 10^{-10}} = \frac{1}{4} = 0.25 = 2.5 \times 10^{-5}

  1. Compare QcQ_c with KcK_c. …
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