Question

Find the rate law for the reaction, $\mathrm{CHCl_3(g) + Cl_2(g) \rightarrow CCl_4(g) + HCl(g)}$ if the order of reaction with respect to $\mathrm{CHCl_3}$ is one and $\frac{1}{2}$ with respect to $\mathrm{Cl_2}$.

• A. Rate $=k[\mathrm{CHCl_3}][\mathrm{Cl_2}]^{1/2}$
• B. Rate $=k[\mathrm{CHCl_3}]^2[\mathrm{Cl_2}]^{1/2}$
• C. Rate $=k[\mathrm{CHCl_3}]^{3/2}[\mathrm{Cl_2}]$
• D. Rate $=k[\mathrm{CHCl_3}]^{1/2}[\mathrm{Cl_2}]$

Hint:

Chemistry Tip: Powers in rate law are experimental orders, not stoichiometric coefficients.

Answer 1

The Correct Option is A

Step 1: General rate law: $$ \text{Rate}=k[A]^m[B]^n $$

Step 2: Here order w.r.t. $\mathrm{CHCl_3}=1$

Step 3: Order w.r.t. $\mathrm{Cl_2}=\frac12$

Step 4: Substitute powers: $$ \text{Rate}=k[\mathrm{CHCl_3}][\mathrm{Cl_2}]^{1/2} $$ $$ \therefore \text{Correct option is (A).} $$