Question

In a reaction \( A + B \to C \), the initial rate of formation of C at 25°C was measured for different initial concentrations of A and B as given. The overall order of the reaction with respect to both A and B is ................. (answer in integer)
% Requires: \usepackage{graphicx} \begin{table}[h] \centering \begin{tabular}{|c|c|c|c|} \hline Experiment & Initial [A] & Initial [B] & Initial rate of formation of C
& (mol L\textsuperscript{-1}) & (mol L\textsuperscript{-1}) & (mol L\textsuperscript{-1} s\textsuperscript{-1})
\hline 1 & 0.4 & 0.3 & 0.078
2 & 0.8 & 0.3 & 0.312
3 & 0.4 & 0.6 & 0.156
4 & 0.8 & 0.6 & 0.624
\hline \end{tabular} \caption{Experimental data showing initial concentrations and rates of formation.} \label{tab:experimental_data} \end{table} \flushleft \newpage

Hint:

To find the orders of reaction, use the rate law and compare experiments where one reactant is held constant while the other is varied. This allows you to isolate the effect of each reactant on the rate.

Answer 1

Step 1: Understanding the reaction rate law. The general form of the rate law for this reaction is: \[ \text{Rate} = k[A]^m[B]^n \] Where:
- \( k \) is the rate constant,
- \( m \) and \( n \) are the orders of the reaction with respect to A and B, respectively.
We are given the experimental data for different concentrations of A and B, and we need to find the orders \( m \) and \( n \).
% Step 2: Using the data from experiments to determine the orders of reaction. From the provided table: \[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & \text{Initial [A] (mol L}^{-1}\text{)} & \text{Initial [B] (mol L}^{-1}\text{)} & \text{Initial rate of formation of C (mol L}^{-1} \text{s}^{-1}\text{)}
\hline 1 & 0.4 & 0.3 & 0.078
2 & 0.8 & 0.3 & 0.312
3 & 0.4 & 0.6 & 0.156
4 & 0.8 & 0.6 & 0.624
\hline \end{array} \] Step 3: Finding the order with respect to A.
We will compare experiments 1 and 2, where the concentration of B is held constant, and only A changes.
For experiments 1 and 2: \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[A_2]^m[B]^n}{k[A_1]^m[B]^n} = \left( \frac{A_2}{A_1} \right)^m \] Substitute the values from experiments 1 and 2: \[ \frac{0.312}{0.078} = \left( \frac{0.8}{0.4} \right)^m \quad \Rightarrow \quad 4 = 2^m \] Solving for \( m \): \[ m = 2 \] Step 4: Finding the order with respect to B.
Now, we compare experiments 1 and 3, where the concentration of A is held constant, and only B changes.
For experiments 1 and 3: \[ \frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k[A]^m[B_3]^n}{k[A]^m[B_1]^n} = \left( \frac{B_3}{B_1} \right)^n \] Substitute the values from experiments 1 and 3: \[ \frac{0.156}{0.078} = \left( \frac{0.6}{0.3} \right)^n \quad \Rightarrow \quad 2 = 2^n \] Solving for \( n \): \[ n = 1 \] Step 5: Determining the overall order.
The overall order of the reaction is the sum of the individual orders: \[ \text{Overall order} = m + n = 2 + 1 = 3 \] Final Answer: \[ \boxed{3} \]