The given rate law for the reaction is:
\(\text{Rate} = k[\text{RCl}]\)
This rate law indicates that the rate of the reaction is directly proportional to the concentration of RCl, which is the alkyl halide.
Now, let's consider the effect of reducing the concentration of the alkyl halide to one-half. Suppose the initial concentration of RCl is \([\text{RCl}]_0\). If we reduce it to \(\frac{1}{2} [\text{RCl}]_0\), the rate of the reaction can be calculated as follows:
\(\text{Rate}' = k \left(\frac{1}{2} [\text{RCl}]_0\right)\)
To compare this rate with the original rate, we can express the original rate as:
\(\text{Rate} = k [\text{RCl}]_0\)
Taking the ratio of the two rates:
\(\frac{\text{Rate}'}{\text{Rate}} = \frac{k \left(\frac{1}{2} [\text{RCl}]_0\right)}{k [\text{RCl}]_0}\)
Canceling out the k term:
\(\frac{\text{Rate}'}{\text{Rate}} = \frac{\left(\frac{1}{2} [\text{RCl}]_0\right)}{[\text{RCl}]_0}\)
Simplifying further:
\(\frac{\text{Rate}'}{\text{Rate}} = \frac{1}{2}\)
This shows that the reduced concentration of the alkyl halide \(\left(\frac{1}{2} [\text{RCl}]_0\right)\) leads to a halving of the rate compared to the original concentration \([\text{RCl}]_0\). Therefore, the rate of the reaction is halved when the concentration of the alkyl halide is reduced to one half.
Hence, option (2) is the correct answer. |
