Question

For the reaction \( A + B \rightarrow P \):
If \( [B] \) is doubled at constant \( [A] \), the rate of reaction doubles.
If \( [A] \) is tripled and \( [B] \) is doubled, the rate of reaction increases by a factor of 6.
Calculate the rate law equation.

Hint:

Reaction order is determined by analyzing how changes in reactant concentration affect the rate of reaction.

Answer 1

The rate law equation is given by: \[ \text{Rate} = k [A]^m [B]^n \] where \( m \) and \( n \) are the reaction orders with respect to \( A \) and \( B \), respectively. 
Step 1: Determine \( n \) From the first condition: \[ [B] \text{ is doubled at constant } [A] \Rightarrow \text{Rate doubles} \] \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left( \frac{[B]_2}{[B]_1} \right)^n \] \[ 2 = (2)^n \] Taking \(\log\): \[ \log 2 = n \log 2 \] \[ n = 1 \] 
Step 2: Determine \( m \) From the second condition: \[ [A] \text{ is tripled and } [B] \text{ is doubled} \Rightarrow \text{Rate increases by a factor of 6} \] \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left( \frac{[A]_2}{[A]_1} \right)^m \times \left( \frac{[B]_2}{[B]_1} \right)^n \] \[ 6 = (3)^m (2)^1 \] \[ 6 = 3^m \times 2 \] \[ \frac{6}{2} = 3^m \] \[ 3^m = 3^1 \] \[ m = 1 \] 
Final Rate Law: \[ \text{Rate} = k [A]^1 [B]^1 \] or \[ \text{Rate} = k [A][B] \]