For a reaction of the type, \( 2X + Y \rightarrow A + B \), the following is the data collected:
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When determining reaction orders, choose experiments where the concentrations of one reactant are varied while others are held constant. This will simplify calculations and give clear values for the exponents in the rate law.
The rate law for the reaction is of the form:
\[
\text{rate} = k[X]^m[Y]^n
\]
Where \( m \) and \( n \) are the orders of reaction with respect to \( X \) and \( Y \), respectively.
We will first determine the value of \( m \) and \( n \) using the method of comparing experiments.
Step 1: Determine \( m \) (Order with respect to X)
From Experiment 1 and Experiment 2, we can compare the rates while keeping \( Y \) constant:
\[
\frac{\text{rate}_2}{\text{rate}_1} = \frac{k[X_2]^m[Y_2]^n}{k[X_1]^m[Y_1]^n} = \frac{(14.4 \times 10^{-2})}{(12.0 \times 10^{-3})} = \left( \frac{0.6}{0.2} \right)^m
\]
\[
\frac{14.4 \times 10^{-2}}{12.0 \times 10^{-3}} = \left( \frac{0.6}{0.2} \right)^m
\]
\[
12 = 3^m
\]
\[
m = 2
\]
Step 2: Determine \( n \) (Order with respect to Y)
Next, we use Experiment 2 and Experiment 3, where \( X \) is kept constant:
\[
\frac{\text{rate}_3}{\text{rate}_2} = \frac{k[X_3]^m[Y_3]^n}{k[X_2]^m[Y_2]^n} = \frac{(5.76 \times 10^{-1})}{(14.4 \times 10^{-2})} = \left( \frac{0.8}{0.4} \right)^n
\]
\[
\frac{5.76 \times 10^{-1}}{14.4 \times 10^{-2}} = \left( \frac{0.8}{0.4} \right)^n
\]
\[
40 = 2^n
\]
\[
n = 3
\]
Step 3: Determine the overall order of the reaction
The overall order of the reaction is the sum of \( m \) and \( n \):
\[
\text{Overall order} = m + n = 2 + 3 = 5
\]
Thus, the overall order of the reaction is 5.
