Question 16 of 69intermediate🔧 ApplyNumerical3 marks

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2ICl(g)⇌I2(g)+Cl2(g);Kc=0.142ICl (g) \rightleftharpoons I_2(g) + Cl_2(g); \quad K_c = 0.14
Correct Answer

At equilibrium, [ICl]=0.446[ICl] = 0.446 M, [I2]=0.167[I_2] = 0.167 M, and [Cl2]=0.167[Cl_2] = 0.167 M.

Exercise: EXERCISES | Q: 6.16 | (Page 42)
For More Understanding

Explanation

To find the equilibrium concentrations, we use an ICE table (Initial, Change, Equilibrium). Let xx be the amount of I2I_2 and Cl2Cl_2 formed. The change in IClICl will be −2x-2x. We substitute these into the equilibrium constant expression Kc=[I2][Cl2][ICl]2K_c = \frac{[I_2][Cl_2]}{[ICl]^2} and solve for xx.

Solution Steps

  1. Initial and Equilibrium Concentrations: For the reaction 2ICl(g)⇌I2(g)+Cl2(g)2ICl (g) \rightleftharpoons I_2(g) + Cl_2(g), let xx be the change in concentration of products. Initial: [ICl]=0.78[ICl] = 0.78 M, [I2]=0[I_2] = 0, [Cl2]=0[Cl_2] = 0 Change: [ICl]=−2x[ICl] = -2x, [I2]=+x[I_2] = +x, [Cl2]=+x[Cl_2] = +x Equilibrium: [ICl]=0.78−2x[ICl] = 0.78 - 2x, [I2]=x[I_2] = x, [Cl2]=x[Cl_2] = x

  2. Equilibrium Constant Expression:

    Kc=[I2][Cl2][ICl]2=0.14K_c = \frac{[I_2][Cl_2]}{[ICl]^2} = 0.14

    Substituting the equilibrium concentrations:

    0.14=(x)(x)(0.78−2x)2=x2(0.78−2x)20.14 = \frac{(x)(x)}{(0.78 - 2x)^2} = \frac{x^2}{(0.78 - 2x)^2}

  3. Solving for xx: Taking the square root of both sides:

    0.14=x0.78−2x\sqrt{0.14} = \frac{x}{0.78 - 2x} 0.374≈x0.78−2x0.374 \approx \frac{x}{0.78 - 2x} 0.374(0.78−2x)=x0.374(0.78 - 2x) = x 0.29172−0.748x=x0.29172 - 0.748x = x 1.748x=0.291721.748x = 0.29172

    x=0.167x = 0.167 M

  4. Calculating Equilibrium Concentrations: [I2]=x=0.167[I_2] = x = 0.167 M [Cl2]=x=0.167[Cl_2] = x = 0.167 M [ICl]=0.78−2x=0.78−2(0.167)=0.446[ICl] = 0.78 - 2x = 0.78 - 2(0.167) = 0.446 M

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