Question

For the reaction: $$ 2A + B \rightarrow 2C + D $$ The following kinetic data were obtained for three different experiments performed at the same temperature: $$ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline \text{I} & 0.10 & 0.10 & 0.10 \\ \text{II} & 0.20 & 0.10 & 0.40 \\ \text{III} & 0.20 & 0.20 & 0.40 \\ \hline \end{array} $$ The total order and order in $[B]$ for the reaction are respectively:

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

Hint:

For zero-order reactions with respect to one reactant, changing its concentration does not affect the reaction rate.

Answer 1

Answer: (E)

The rate law for the reaction is: \[ {rate} = k[A]^m[B]^n \] where \(m\) is the order with respect to \(A\) and \(n\) is the order with respect to \(B\). 
Step 1: Comparing experiments I and II, we see that the concentration of \(B\) is constant, while the concentration of \(A\) doubles. 
The rate also quadruples, indicating that the reaction is second order with respect to \(A\), so \(m = 2\). \[ \frac{{rate}_{II}}{{rate}_{I}} = \frac{k(0.20)^m(0.10)^n}{k(0.10)^m(0.10)^n} = \frac{0.40}{0.10} = 4 \] \[ \Rightarrow \left( \frac{0.20}{0.10} \right)^m = 4 \quad \Rightarrow \quad m = 2 \] 
Step 2: Now comparing experiments II and III, we see that the concentration of \(A\) is constant while the concentration of \(B\) doubles. 
The rate does not change, indicating that the reaction is zero order with respect to \(B\), so \(n = 0\). \[ \frac{{rate}_{III}}{{rate}_{II}} = \frac{k(0.20)^m(0.20)^n}{k(0.20)^m(0.10)^n} = \frac{0.40}{0.40} = 1 \] \[ \Rightarrow \left( \frac{0.20}{0.10} \right)^n = 1 \quad \Rightarrow \quad n = 0 \] 
Thus, the total order of the reaction is \(m + n = 2 + 0 = 2\) and the order with respect to \(B\) is \(n = 0\).