Question 21 of 69intermediate🔧 ApplyNumerical3 marks

Equilibrium constant, KcK_c for the reaction

N2(g)+3H2(g)⇌2NH3(g) at 500 K is 0.061N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \text{ at 500 K is 0.061}

At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L−1^{-1} N2_2, 2.0 mol L−1^{-1} H2_2 and 0.5 mol L−1^{-1} NH3_3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Correct Answer

The reaction is not at equilibrium and will proceed in the forward direction to reach equilibrium.

Exercise: EXERCISES | Q: 6.21 | (Page 43)
For More Understanding

Explanation

To determine if the reaction is at equilibrium, we calculate the reaction quotient QcQ_c using the given concentrations and compare it with the equilibrium constant KcK_c. If Qc<KcQ_c < K_c, the reaction proceeds in the forward direction to form more products until equilibrium is reached.

Solution Steps

  1. Step 1: Write the expression for the reaction quotient QcQ_c.

    Qc=[NH3]2[N2][H2]3Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}

  2. Step 2: Substitute the given molar concentrations into the expression. [N2]=3.0[N_2] = 3.0 M, [H2]=2.0[H_2] = 2.0 M, [NH3]=0.5[NH_3] = 0.5 M

    Qc=(0.5)2(3.0)×(2.0)3=0.253.0×8.0=0.2524.0≈0.0104Q_c = \frac{(0.5)^2}{(3.0) \times (2.0)^3} = \frac{0.25}{3.0 \times 8.0} = \frac{0.25}{24.0} \approx 0.0104

  3. Step 3: Compare QcQ_c with the given KcK_c. Given Kc=0.061K_c = 0.061. Since QcQ_c (0.01040.0104) is less than KcK_c (0.0610.061), the reaction is not at equilibrium.

  4. Step 4: Determine the direction of the reaction. Because Qc<KcQ_c < K_c, the reaction will proceed in the forward direction to increase the concentration of NH3NH_3 and decrease the concentrations of N2N_2 and H2H_2 until equilibrium is established.

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