Rate law for the reaction, `A + 2B to C` is found to be
Rate `= k [A] [B]`
Concentration of reactant 'B' is doubled keeping the concentration of 'A' constant, the value of rate constant will be _______

Options

A. the same
B. doubled
C. quadrupled
D. halved

Explanation

To solve the problem step by step, we will analyze the given reaction and the effect of changing the concentration of reactant B on the rate of the reaction.

### Step 1: Write the rate law for the reaction
The given reaction is:
\[ A + 2B \rightarrow C \]

The rate law is provided as:
\[ \text{Rate} = k [A]^1 [B]^1 \]
This indicates that the rate of the reaction depends on the concentration of A and B, both raised to the power of 1.

### Step 2: Define the initial rate
Let’s denote the initial rate of the reaction as \( R_1 \):
\[ R_1 = k [A] [B] \]

### Step 3: Change the concentration of B
According to the problem, the concentration of reactant B is doubled while keeping the concentration of A constant. Therefore, if the initial concentration of B is \( [B] \), the new concentration of B, denoted as \( [B'] \), will be:
\[ [B'] = 2[B] \]

### Step 4: Write the new rate law with the changed concentration
The new rate, denoted as \( R_2 \), can be expressed as:
\[ R_2 = k [A] [B'] \]
Substituting the new concentration of B:
\[ R_2 = k [A] (2[B]) \]
\[ R_2 = 2k [A] [B] \]

### Step 5: Relate the new rate to the initial rate
From the expression of \( R_2 \), we can see that:
\[ R_2 = 2R_1 \]
This means that the new rate \( R_2 \) is twice the initial rate \( R_1 \).

### Step 6: Conclusion about the rate constant
The rate constant \( k \) remains unchanged regardless of the concentrations of the reactants. Therefore, even though the rate of the reaction has doubled, the value of the rate constant \( k \) does not change.

### Final Answer
Thus, the value of the rate constant remains the same:
\[ \text{Value of rate constant } k = k \]

---