Question

In the reaction, \(A\rightarrow\) products, if the concentration of the reactant is doubled, rate of the reaction remains unchanged. The order of the reaction with respect to \(A\) is

• A. 1
• B. 2
• C. 0.5
• D. 0

Hint:

For a zero-order reaction, \[ \text{Rate}=k[A]^0=k \] So the rate remains unchanged even if the concentration of the reactant is increased or decreased.

Answer 1

The Correct Option is D

Step 1: Write the rate law.
For the reaction \[ A\rightarrow \text{products} \] let the rate law be \[ \text{Rate}=k[A]^n \] where \(n\) is the order of reaction with respect to \(A\).

Step 2: Apply the condition given in the question.
When concentration of \(A\) is doubled, \[ [A]\rightarrow 2[A] \] The new rate becomes \[ \text{New Rate}=k(2[A])^n \] \[ =k2^n[A]^n \] \[ =2^n \times \text{Original Rate} \]

Step 3: Use the fact that rate remains unchanged.
According to the question, doubling the concentration does not change the rate. Therefore, \[ 2^n=1 \] This is possible only when \[ n=0 \]

Step 4: Interpret the result.
If the order is zero, the rate law becomes \[ \text{Rate}=k[A]^0 \] Since, \[ [A]^0=1 \] we get \[ \text{Rate}=k \] Thus, the rate is independent of the concentration of \(A\).

Step 5: Final conclusion.
Therefore, the order of reaction with respect to \(A\) is \[ \boxed{0} \] Hence, option (4) is correct.