Question

For a reaction \( A + B \to \text{products} \), the rate of the reaction was doubled when the concentration of \( A \) was doubled. When the concentrations of \( A \) and \( B \) were doubled, the rate was again doubled. The order of the reaction with respect to \( A \) and \( B \) are:

• A. 1, 1
• B. 2, 0
• C. 1, 0
• D. 0, 1

Hint:

If doubling a reactant's concentration results in the rate doubling, it's 1st order. If it results in the rate quadrupling, it's 2nd order. If the rate doesn't change at all, it's 0th order.

Answer 1

The Correct Option is C

Concept: The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants.
• Rate law: \( \text{Rate} = k[A]^x [B]^y \)
• \( x \): Order with respect to reactant \( A \).
• \( y \): Order with respect to reactant \( B \).
• We can determine these orders by observing how changes in concentration affect the initial rate.

Step 1: Finding the order with respect to \( A \).
The problem states: Rate doubles when \([A]\) is doubled (keeping \([B]\) constant). \[ 2 \times \text{Rate} = k(2[A])^x [B]^y \] Dividing this by the original rate equation: \[ 2 = 2^x \quad \Rightarrow \quad \mathbf{x = 1} \] So, the reaction is first order with respect to \( A \).

Step 2: Finding the order with respect to \( B \).
The problem states: When both \([A]\) and \([B]\) are doubled, the rate is again doubled (compared to the original rate). \[ 2 \times \text{Rate} = k(2[A])^1 (2[B])^y \] Since we already know from Step 1 that doubling \([A]\) alone doubles the rate, the doubling of \([B]\) must have had no effect on the rate. \[ 2 = 2^1 \cdot 2^y \] \[ 2 = 2 \cdot 2^y \quad \Rightarrow \quad 1 = 2^y \quad \Rightarrow \quad \mathbf{y = 0} \] So, the reaction is zero order with respect to \( B \).

Step 3: Conclusion.
The order with respect to \( A \) is 1 and with respect to \( B \) is 0. This matches option (C).