The rate law for the reaction
`RCl + NaOH(aq) rarr ROH + NaCl` is given by
Rate `= k[RCl]`. The rate of the reaction will be

Options

A. Doubled on doubling the concentration of NaOH
B. Halved on reducing the concentration of alkyl halide to one half
C. Decreases on increasing the temperature
D. Unaffected by increasing the temperature

Answer

To solve the problem regarding the rate law for the reaction \( RCl + NaOH(aq) \rightarrow ROH + NaCl \), we need to analyze the given information step by step.

### Step 1: Understanding the Rate Law
The rate law for the reaction is given as:
\[ \text{Rate} = k[RCl] \]
where \( k \) is the rate constant and \([RCl]\) is the concentration of the reactant \( RCl \).

### Step 2: Analyzing the Dependence of Rate on Concentration
From the rate law, we can see that the rate of the reaction is directly proportional to the concentration of \( RCl \). This means that if the concentration of \( RCl \) increases, the rate of the reaction will also increase.

### Step 3: Considering Changes in Concentration
If we were to double the concentration of \( RCl \), the rate would also double. Conversely, if we were to reduce the concentration of \( RCl \) to half, the rate would also be halved.

### Step 4: Evaluating the Role of Other Reactants
The rate law provided does not include \( NaOH \), which indicates that the rate of the reaction is independent of the concentration of \( NaOH \). Therefore, changes in the concentration of \( NaOH \) will not affect the rate of the reaction.

### Step 5: Conclusion on the Rate of Reaction
Based on the analysis, we can conclude that:
1. The rate of the reaction is directly proportional to the concentration of \( RCl \).
2. The rate does not depend on the concentration of \( NaOH \).

Thus, if we reduce the concentration of \( RCl \) to half, the rate of the reaction will also become half.

### Final Answer
The rate of the reaction will depend on the concentration of \( RCl \) and will be halved if the concentration of \( RCl \) is halved.

---