Question

Consider the following reaction: CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) The initial rate of the reaction is measured at several different concentrations of the reactants with the following results: [CHCl3] (M) [Cl2] (M) Initial rate (M/s) 0.010 0.010 0.0035 0.020 0.010 0.0069 0.020 0.020 0.0098 0.040 0.040 0.027 From the data, choose the correct rate law for the reaction. rate=k[CHCl3][Cl2]2 rate=k[CHCl3][Cl2]12 rate=k[CHCl3]2[Cl2] rate=k[CHCl3]12[Cl2]

Answer 1

Answer : The correct rate law for the reaction is,

$\text{Rate}=k[CHCl_3][Cl_2]^{1/2}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$CHCl_3(g)+Cl_2(g)\rightarrow CCl_2(g)+HCl(g)$

Rate law expression for the reaction:

$\text{Rate}=k[CHCl_3]^a[Cl_2]^b$

where,

a = order with respect to $CHCl_3$

b = order with respect to $Cl_2$

Expression for rate law for first observation:

$0.0035=k(0.010)^a(0.010)^b$ ....(1)

Expression for rate law for second observation:

$0.0069=k(0.020)^a(0.010)^b$ ....(2)

Expression for rate law for third observation:

$0.0098=k(0.020)^a(0.020)^b$ ....(3)

Expression for rate law for fourth observation:

$0.027=k(0.040)^a(0.040)^b$ ....(4)

Dividing 1 from 2, we get:

$\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1$

Dividing 2 from 3, we get:

$\frac{0.0098}{0.0069}=\frac{k(0.020)^a(0.020)^b}{k(0.020)^a(0.010)^b}\\\\1.42=2^b\\b=\frac{1}{2}$

Calculation used :

$1.42=2^b\\\log (1.42)=b\log 2\\\log (\frac{1.42}{2})=b\\b=0.5=\frac{1}{2}$

Thus, the rate law becomes:

$\text{Rate}=k[CHCl_3]^1[Cl_2]^{1/2}$