Question

For the reaction \(A \longrightarrow\) products the following data are obtained. $$\begin{array}{cll} \hline {\text { Experiment 1 }} & &{\text { Experiment 2 }} \\ \hline [\mathrm{A}]=1.204 \mathrm{M} & t=0 \mathrm{min} & {[\mathrm{A}]=2.408 \mathrm{M}} & t=0 \mathrm{min}\\\ {[\mathrm{A}]=1.180 \mathrm{M}} & t=1.0 \mathrm{min} & {[\mathrm{A}]=?} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=0.602 \mathrm{M}} & t=35 \mathrm{min} & {[\mathrm{A}]=?} & t=30 \mathrm{min} \\ \hline \end{array}$$ (a) Determine the initial rate of reaction in Experiment 1. (b) If the reaction is second order, what will be \([\mathrm{A}]\) at \(t=1.0\) min in Experiment 2? (c) If the reaction is first order, what will be \([\mathrm{A}]\) at 30 min in Experiment 2?

Short answer

(a) The initial rate of reaction is 0.024 M/min. (b) If the reaction is second order, [A] at t=1.0 min in Experiment 2 is approximately 0.962 M. (c) If the reaction is first order, [A] at t=30 min in Experiment 2 is approximately 3.215 M.

Step-by-step solution

Determine the initial rate of reaction

From the data given, at \(t=0\) min, we have \([A]=1.204\) M and at \(t=1.0\) min, \([A]=1.180\) M. The rate of reaction is defined as the change in concentration of reactant A over the change in time. So, \(Rate=-\Delta[A]/\Delta t\). Substituting the values in, we have \(- (1.180-1.204)/(1.0-0)\) which gives \(Rate = 0.024\) M/min.

Calculate [A] at t=1.0 min in Experiment 2 given a second order reaction

For a second order reaction, the rate law is \(\frac{1}{[A]_t}\) = \(\frac{1}{[A]_0}\) + kt. Using the rate calculated in step 1 as k, and considering the initial concentration of A in Experiment 2 as 2.408 M, we can calculate the concentration of A at 1.0 min. Rearranging the second order rate law gives \([A]_t = \frac{1} {\frac{1}{2.408} + 0.024*1}\), which evaluates to approximately 0.962 M.

Calculate [A] at t=30 min in Experiment 2 given a first order reaction

For a first order reaction, the rate law is \(\ln[A]_t = -kt + \ln[A]_0\). Using the rate calculated in step 1 as k, and the initial concentration of A in Experiment 2, we can calculate the concentration of A at 30 min. Substituting the values into the rate law gives \( \ln[A]_{30} = -0.024*30 + \ln[2.408]\), which evaluates to approximately 1.167. Exponentiating this value gives \([A]_{30} = e^{1.167}\), which evaluates to approximately 3.215 M.

Key concepts

Initial Rate of Reaction

When studying chemical kinetics, the initial rate of a reaction is a crucial measure that helps us understand how quickly a reaction begins. It is the rate of change in the concentration of a reactant, right at the start, when time (\(t\)) is zero. This is especially useful because it avoids any complications from secondary reactions or product buildup.
At the beginning of the reaction, we measure how quickly the concentration of a reactant, like \([A]\), decreases. In Experiment 1, we have the concentrations at \(t=0\) min (\([A] = 1.204\) M) and \(t=1.0\) min (\([A] = 1.180\) M).
To find the initial rate, use the formula: \[ \text{Rate} = -\frac{\Delta [A]}{\Delta t}\]Here, \(\Delta [A]\) is the change in concentration, and \(\Delta t\) is the change in time. Substituting the values gives:- \((1.180 - 1.204) / (1.0 - 0)\), resulting in an initial rate of \(0.024\) M/min.
This tells us how fast the reaction is proceeding right from the start.

First Order Reaction

In first order reactions, the rate depends on the concentration of just one reactant. These reactions follow the simple rate law:\[ \ln[A]_t = -kt + \ln[A]_0 \]where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and k is the rate constant. These equations are handy because they let us predict how concentration changes over time.
Consider Experiment 2. To find \([A]\) at \(t=30\) min, given \([A]_0 = 2.408\) M and assuming it's a first order reaction, substitute into the rate law:
\[ \ln[A]_{30} = -0.024 \times 30 + \ln[2.408]\]
After calculating, we find the natural logarithm of \([A]_{30}\) to be approximately \(1.167\).
To find \([A]\), convert the natural log value back using the exponential function: \([A]_{30} = e^{1.167}\), which is roughly \(3.215\) M.
This implies that, throughout time, the concentration reduces at a rate proportional to its current value.

Second Order Reaction

Second order reactions depend on the concentration of one reactant squared or on two different reactants. The rate law for these is:\[ \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \]This equation helps track the concentration changes over time. For Experiment 2, we need to find \([A]\) at \(t=1.0\) min, starting with \([A]_0 = 2.408\) M.
Using the second order rate law and assuming \(k = 0.024\) M/min,\[ \frac{1}{[A]_t} = \frac{1}{2.408} + 0.024 \times 1\]Rearranging gives \([A]_t = \frac{1}{\frac{1}{2.408} + 0.024}\). This calculates to approximately \(0.962\) M.
This lower concentration shows how the reaction proceeds, emphasizing its greater reliance on initial concentration compared to first order kinetics.