Questions: The following data was collected at room temperature for the reaction: 2 A(aq) + B(aq) → 3 C(aq) Trial [A](M) [B](M) Initial Rate (M / s) 1 0.1 0.1 2 2 0.3 0.3 6 3 0.1 0.3 2 4 0.2 0.4 4 What is the overall order of the reaction? a. 0th b. 1st c. 2nd d. 3rd e. Can't tell from this information alone

Solution Steps

The rate law for the reaction \(2A (aq) + B (aq) \rightarrow 3C (aq)\) can be written as: \[ \text{Rate} = k[A]^m[B]^n \] where \(k\) is the rate constant, and \(m\) and \(n\) are the orders of the reaction with respect to \(A\) and \(B\), respectively.

To find the order with respect to \(A\), compare trials where the concentration of \(B\) is constant. Compare trials 1 and 2: \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[A]_2^m[B]_2^n}{k[A]_1^m[B]_1^n} \] \[ \frac{6}{2} = \frac{[0.3]^m[0.1]^n}{[0.1]^m[0.1]^n} \] \[ 3 = \left(\frac{0.3}{0.1}\right)^m \] \[ 3 = 3^m \] \[ m = 1 \]

To find the order with respect to \(B\), compare trials where the concentration of \(A\) is constant. Compare trials 3 and 4: \[ \frac{\text{Rate}_4}{\text{Rate}_3} = \frac{k[A]_4^m[B]_4^n}{k[A]_3^m[B]_3^n} \] \[ \frac{4}{2} = \frac{[0.2]^m[0.4]^n}{[0.2]^m[0.3]^n} \] \[ 2 = \left(\frac{0.4}{0.3}\right)^n \] \[ 2 = \left(\frac{4}{3}\right)^n \] Taking the natural logarithm on both sides: \[ \ln(2) = n \ln\left(\frac{4}{3}\right) \] \[ n = \frac{\ln(2)}{\ln\left(\frac{4}{3}\right)} \approx 2 \]

Final Answer