For the reaction \( \ce{CHCl3 + Cl2 -> CCl4 + HCl} \), the experimental rate law is Rate = k[CHCl3][Cl2]1/2. At a certain temperature, the rate is 4.0 × 10−4 mol L−1 s−1 when [CHCl3] = 0.20 mol L−1 and [Cl2] = 0.16 mol L−1. What is the rate at the same temperature when [CHCl3] = 0.10 mol L−1 and [Cl2] = 0.64 mol L−1?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4.0 × 10−4 mol L−1 s−1
In this rate law [CHCl3] has the power one and [Cl2] has the power one-half. At the same temperature k does not change, so the ratio of the two rates depends only on the concentration terms.
\( \displaystyle \frac{r_2}{r_1} = \frac{[\ce{CHCl3}]_2}{[\ce{CHCl3}]_1} \times \left(\frac{[\ce{Cl2}]_2}{[\ce{Cl2}]_1}\right)^{1/2} \)
\( \displaystyle \frac{r_2}{r_1} = \frac{0.10}{0.20} \times \left(\frac{0.64}{0.16}\right)^{1/2} = \frac{1}{2} \times \sqrt{4} = \frac{1}{2} \times 2 = 1 \)
\( \displaystyle r_2 = r_1 = 4.0 \times 10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}} \)
Halving [CHCl3] halves the rate, and making [Cl2] four times raises the rate only by \( \sqrt{4} = 2 \). The two effects cancel, so the rate stays the same.
NCERT (2026-27 edition):
3.2.2 Rate Expression and Rate Constant – “Thus, rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation.” … “Some other examples are given below:”
\( \displaystyle \ce{CHCl3 + Cl2 -> CCl4 + HCl} \qquad \text{Rate} = k[\ce{CHCl3}][\ce{Cl2}]^{1/2} \)