Question 22 of 69intermediate🔧 ApplyNumerical3 marks

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2BrCl(g)⇌Br2(g)+Cl2(g)2BrCl (g) \rightleftharpoons Br_2(g) + Cl_2(g)

for which Kc=32K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3×10−33.3 \times 10^{-3} mol L−1^{-1}, what is its molar concentration in the mixture at equilibrium?

Correct Answer

The molar concentration of BrCl in the mixture at equilibrium is 0.268×10−30.268 \times 10^{-3} mol L−1^{-1}.

Exercise: EXERCISES | Q: 6.22 | (Page 43)
For More Understanding

Explanation

To find the equilibrium concentration of BrCl, we set up an ICE (Initial, Change, Equilibrium) table. Since the reaction starts with pure BrCl, the initial concentrations of the products are zero. As the reaction proceeds to equilibrium, 2x2x moles of BrCl decompose to form xx moles each of Br2Br_2 and Cl2Cl_2. By substituting these equilibrium concentrations into the KcK_c expression and taking the square root of both sides, we can easily solve for xx and subsequently find the equilibrium concentration of BrCl.

Solution Steps

  1. Step 1: Write the balanced equation and initial concentrations.

    2BrCl(g)⇌Br2(g)+Cl2(g)2BrCl(g) \rightleftharpoons Br_2(g) + Cl_2(g)

    Initial concentration: 3.3×10−33.3 \times 10^{-3}, 00, 00

  2. Step 2: Let xx be the amount of Br2Br_2 and Cl2Cl_2 formed at equilibrium. The change in BrClBrCl will be −2x-2x. Equilibrium concentration: 3.3×10−3−2x3.3 \times 10^{-3} - 2x, xx, xx

  3. Step 3: Write the expression for KcK_c and substitute the equilibrium concentrations.

    Kc=[Br2][Cl2][BrCl]2K_c = \frac{[Br_2][Cl_2]}{[BrCl]^2} 32=x×x(3.3×10−3−2x)232 = \frac{x \times x}{(3.3 \times 10^{-3} - 2x)^2}

  4. Step 4: Take the square root of both sides to solve for xx.

    32=x3.3×10−3−2x\sqrt{32} = \frac{x}{3.3 \times 10^{-3} - 2x} 5.656=x3.3×10−3−2x5.656 = \frac{x}{3.3 \times 10^{-3} - 2x}

  5. Step 5: Rearrange to find xx.

    5.656×(3.3×10−3−2x)=x5.656 \times (3.3 \times 10^{-3} - 2x) = x 18.66×10−3−11.312x=x18.66 \times 10^{-3} - 11.312x = x 18.66×10−3=12.312x18.66 \times 10^{-3} = 12.312x

    x=18.66×10−312.312=1.516×10−3x = \frac{18.66 \times 10^{-3}}{12.312} = 1.516 \times 10^{-3} M

  6. Step 6: Calculate the equilibrium concentration of BrClBrCl.

    [BrCl]=3.3×10−3−2x[BrCl] = 3.3 \times 10^{-3} - 2x [BrCl]=3.3×10−3−2(1.516×10−3)[BrCl] = 3.3 \times 10^{-3} - 2(1.516 \times 10^{-3})

    [BrCl]=3.3×10−3−3.032×10−3=0.268×10−3[BrCl] = 3.3 \times 10^{-3} - 3.032 \times 10^{-3} = 0.268 \times 10^{-3} M

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