Question 19 of 69intermediate🔧 ApplyNumerical3 marks

A sample of pure PCl5_5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5_5 was found to be 0.5×10−10.5 \times 10^{-1} mol L−1^{-1}. If value of KcK_c is 8.3×10−38.3 \times 10^{-3}, what are the concentrations of PCl3_3 and Cl2_2 at equilibrium?

PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)
Correct Answer

The concentrations of PCl3PCl_3 and Cl2Cl_2 at equilibrium are 2.04×10−22.04 \times 10^{-2} mol L−1^{-1} each.

Exercise: EXERCISES | Q: 6.19 | (Page 43)
For More Understanding

Explanation

For the dissociation of PCl5PCl_5 in an evacuated vessel, the initial concentrations of the products are zero. At equilibrium, the concentrations of PCl3PCl_3 and Cl2Cl_2 will be equal since they are produced in a 1:1 molar ratio. By substituting the given equilibrium concentration of PCl5PCl_5 and the value of KcK_c into the equilibrium constant expression, we can solve for the concentration of the products.

Solution Steps

  1. Step 1: Write the equilibrium constant expression for the reaction PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g).

    Kc=[PCl3][Cl2][PCl5]K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}

  2. Step 2: Let the equilibrium concentrations of PCl3PCl_3 and Cl2Cl_2 be xx. Since the vessel was initially evacuated, [PCl3]=[Cl2]=x[PCl_3] = [Cl_2] = x at equilibrium.

  3. Step 3: Substitute the given values into the expression. Given [PCl5]=0.5×10−1[PCl_5] = 0.5 \times 10^{-1} mol L−1^{-1} and Kc=8.3×10−3K_c = 8.3 \times 10^{-3}.

    8.3×10−3=x×x0.5×10−1=x20.058.3 \times 10^{-3} = \frac{x \times x}{0.5 \times 10^{-1}} = \frac{x^2}{0.05}

  4. Step 4: Solve for xx.

    x2=8.3×10−3×0.05=41.5×10−5=4.15×10−4x^2 = 8.3 \times 10^{-3} \times 0.05 = 41.5 \times 10^{-5} = 4.15 \times 10^{-4}

    x=4.15×10−4=2.037×10−2x = \sqrt{4.15 \times 10^{-4}} = 2.037 \times 10^{-2} mol L−1^{-1}

  5. Step 5: State the final concentrations. [PCl3]=[Cl2]=2.04×10−2[PCl_3] = [Cl_2] = 2.04 \times 10^{-2} mol L−1^{-1}

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