Question

The rate of a reaction:
A + B −→ product
is given below as a function of different initial concentrations of A and B.

Experiment	\([A]\) (mol L\(^{-1}\))	\([B]\) (mol L\(^{-1}\))	Initial Rate (mol L\(^{-1}\) min\(^{-1}\))
1	0.01	0.01	\(5 \times 10^{-3}\)
2	0.02	0.01	\(1 \times 10^{-2}\)
3	0.01	0.02	\(5 \times 10^{-3}\)

Hint:

When determining the order of reaction, compare experiments where one concentration is kept constant and use the rate law to solve for the reaction order.

Answer 1

The rate law for the reaction can be written as: \[ \text{Rate} = k[\text{A}]^m[\text{B}]^n \] where \( k \) is the rate constant, \( m \) is the order with respect to A, and \( n \) is the order with respect to B. From Experiment 1 and Experiment 2, where the concentration of B is constant, we can determine the order with respect to A: \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left( \frac{[\text{A}]_2}{[\text{A}]_1} \right)^m \] Substituting the known values: \[ \frac{1 \times 10^{-2}}{5 \times 10^{-3}} = \left( \frac{0.02}{0.01} \right)^m \Rightarrow 2 = 2^m \] Thus, \( m = 1 \). From Experiment 1 and Experiment 3, where the concentration of A is constant, we can determine the order with respect to B: \[ \frac{\text{Rate}_3}{\text{Rate}_1} = \left( \frac{[\text{B}]_3}{[\text{B}]_1} \right)^n \] Substituting the known values: \[ \frac{5 \times 10^{-3}}{5 \times 10^{-3}} = \left( \frac{0.02}{0.01} \right)^n \Rightarrow 1 = 2^n \] Thus, \( n = 0 \). The rate law is: \[ \text{Rate} = k[\text{A}]^1[\text{B}]^0 = k[\text{A}] \] To calculate the rate constant, use the data from any experiment, say Experiment 1: \[ 5 \times 10^{-3} = k(0.01) \Rightarrow k = \frac{5 \times 10^{-3}}{0.01} = 0.5 \, \text{L}^{-1} \text{min}^{-1} \]