Question

When the concentration of the reactant in a given reaction is halved and if the rate of reaction is halved, the order of the reaction is:

• A. 3
• B. 0
• C. 2
• D. 1

Hint:

For first-order reactions: \[ \text{Rate} \propto \text{Concentration} \] So halving concentration halves the rate.

Answer 1

The Correct Option is D

Concept: The rate law for a reaction is: \[ r = k[A]^n \] where:
• \(r\) = rate of reaction
• \(k\) = rate constant
• \([A]\) = concentration of reactant
• \(n\) = order of reaction

Step 1: Writing initial rate equation.
Suppose the initial concentration is \(a\). Then: \[ r_1 = ka^n \]

Step 2: Applying the changed condition.
According to the question: \[ [A]_2 = \frac{a}{2} \] Rate is also halved: \[ r_2 = \frac{r_1}{2} \] Therefore: \[ \frac{r_1}{2} = k\left(\frac{a}{2}\right)^n \]

Step 3: Dividing equations.
Divide second equation by first equation: \[ \frac{r_1/2}{r_1} = \frac{k(a/2)^n}{ka^n} \] \[ \frac{1}{2} = \left(\frac{1}{2}\right)^n \] Comparing powers: \[ n = 1 \] Thus, the reaction is first order.