Question

Consider the following reaction
\[A + B \rightarrow C\]
The time taken for A to become 1/4th of its initial concentration is twice the time taken to become 1/2 of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis.
The overall order of the reaction is ____.

Answer 1

Correct Answer: 1

To determine the overall order of the reaction \(A + B \rightarrow C\), we analyze the kinetic behavior described. First, we consider the information that the time for \(A\) to become \( \frac{1}{4} \) of its initial concentration is twice the time for it to become \( \frac{1}{2} \). This characteristic is indicative of a first-order reaction since the integrated rate law for a first-order reaction can be expressed as:
\[ [A] = [A]_0 e^{-kt} \]
For a first-order reaction, the time (\(t\)) to reach a fraction of the concentration is given by:
\[ t = \frac{\ln(\text{fraction})}{k} \]
Let's consider the times \(t_1\) and \(t_2\):
\* Time \(t_1\) for \(A\) to become \( \frac{1}{2} \times [A]_0\):
\[ t_1 = \frac{\ln(2)}{k} \]
\* Time \(t_2\) for \(A\) to become \( \frac{1}{4} \times [A]_0\):
\[ t_2 = \frac{\ln(4)}{k} \]
Given \(t_2 = 2t_1\), substitute:
\[ \frac{\ln(4)}{k} = 2 \times \frac{\ln(2)}{k} \]
Solving confirms:
\[ 2\ln(2) = 2\ln(2) \]
This behavior confirms the reaction's dependence on first-order kinetics for \(A\). Next, consider the graph of the change in concentration of \(B\) versus time, giving a straight line with a negative slope. This suggests zero-order kinetics concerning \(B\) since a zero-order reaction displays a linear decrease in concentration over time as per the equation:
\[ [B] = [B]_0 - kt \]
Thus, the overall order is the sum of the orders with respect to \(A\) and \(B\):
\[ \text{Order of } [A] = 1, \quad \text{Order of } [B] = 0 \]
Overall reaction order = \( 1 \).
The determined order fits the provided range (1,1), consistent with first-order behavior.

Answer 2

Order with respect to A
For a first-order reaction:
\[t_{75\%} = 2 \times t_{50\%}.\]
This is consistent with the information given, so the reaction is first order with respect to A.
Order with respect to B The plot of [B] versus $t$ is a straight line, which indicates that the reaction is zero order with respect to B.
Overall order of the reaction:
\[\text{Order} = 1 \, (\text{w.r.t. A}) + 0 \, (\text{w.r.t. B}) = 1.\]
Final Answer:\[1.\]