Question

Consider the two different first-order reactions given below: \[\text{A + B} \rightarrow \text{C (Reaction 1)} \\\text{P} \rightarrow \text{Q (Reaction 2)}\]The ratio of the half-life of Reaction 1 : Reaction 2 is $5 : 2$. If $t_1$ and $t_2$ represent the time taken to complete $\frac{2}{3}^\text{rd}$ and $\frac{4}{5}^\text{th}$ of Reaction 1 and Reaction 2, respectively, then the value of the ratio $t_1 : t_2$ is ____ $\times 10^{-1}$ (nearest integer).[Given: $\log_{10}(3) = 0.477$ and $\log_{10}(5) = 0.699$]

Answer 1

Correct Answer: 17

The problem asks for the ratio of time \(t_1\) to \(t_2\), where \(t_1\) is the time for \(\frac{2}{3}\) completion of a first-order reaction (Reaction 1) and \(t_2\) is the time for \(\frac{4}{5}\) completion of another first-order reaction (Reaction 2). We are given the ratio of their half-lives.

Concept Used:

For a first-order reaction, the integrated rate law is given by:

\[ k = \frac{1}{t} \ln\left(\frac{[A_0]}{[A_t]}\right) \]

where \(k\) is the rate constant, \(t\) is time, \([A_0]\) is the initial concentration, and \([A_t]\) is the concentration at time \(t\). This can also be written using base-10 logarithm as:

\[ k = \frac{2.303}{t} \log_{10}\left(\frac{[A_0]}{[A_t]}\right) \]

The half-life (\(t_{1/2}\)) of a first-order reaction is related to the rate constant by:

\[ t_{1/2} = \frac{\ln(2)}{k} \]

From this, it is clear that the rate constant \(k\) is inversely proportional to the half-life \(t_{1/2}\).

Step-by-Step Solution:

Step 1: Relate the rate constants of the two reactions using the given half-life ratio.

Let the rate constants for Reaction 1 and Reaction 2 be \(k_1\) and \(k_2\), and their half-lives be \((t_{1/2})_1\) and \((t_{1/2})_2\), respectively. Since \(k \propto 1/t_{1/2}\), the ratio of the rate constants is the inverse of the ratio of their half-lives.

Given:

\[ \frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{5}{2} \]

Therefore, the ratio of the rate constants is:

\[ \frac{k_1}{k_2} = \frac{(t_{1/2})_2}{(t_{1/2})_1} = \frac{2}{5} \]

Step 2: Calculate the expression for \(t_1\) for Reaction 1.

Reaction 1 is \(\frac{2}{3}\) complete, which means the fraction of reactant remaining is \(1 - \frac{2}{3} = \frac{1}{3}\). So, \(\frac{[A_t]}{[A_0]} = \frac{1}{3}\), or \(\frac{[A_0]}{[A_t]} = 3\).

Using the integrated rate law for Reaction 1:

\[ k_1 = \frac{1}{t_1} \ln(3) \implies t_1 = \frac{\ln(3)}{k_1} \]

Step 3: Calculate the expression for \(t_2\) for Reaction 2.

Reaction 2 is \(\frac{4}{5}\) complete, which means the fraction of reactant remaining is \(1 - \frac{4}{5} = \frac{1}{5}\). So, \(\frac{[P_t]}{[P_0]} = \frac{1}{5}\), or \(\frac{[P_0]}{[P_t]} = 5\).

Using the integrated rate law for Reaction 2:

\[ k_2 = \frac{1}{t_2} \ln(5) \implies t_2 = \frac{\ln(5)}{k_2} \]

Step 4: Determine the ratio \(t_1 : t_2\).

Now, we can find the ratio of \(t_1\) to \(t_2\):

\[ \frac{t_1}{t_2} = \frac{\ln(3)/k_1}{\ln(5)/k_2} = \frac{\ln(3)}{\ln(5)} \times \frac{k_2}{k_1} \]

From Step 1, we know \(\frac{k_1}{k_2} = \frac{2}{5}\), which implies \(\frac{k_2}{k_1} = \frac{5}{2}\). Substituting this into the equation:

\[ \frac{t_1}{t_2} = \frac{\ln(3)}{\ln(5)} \times \frac{5}{2} \]

Final Computation & Result:

Step 5: Substitute the given log values and calculate the final ratio.

The ratio of natural logarithms is equal to the ratio of base-10 logarithms:

\[ \frac{\ln(3)}{\ln(5)} = \frac{2.303 \log_{10}(3)}{2.303 \log_{10}(5)} = \frac{\log_{10}(3)}{\log_{10}(5)} \]

Given \(\log_{10}(3) = 0.477\) and \(\log_{10}(5) = 0.699\):

\[ \frac{t_1}{t_2} = \frac{0.477}{0.699} \times \frac{5}{2} \] \[ \frac{t_1}{t_2} \approx 0.6824 \times 2.5 \approx 1.706 \]

The problem asks for the value of the ratio to be expressed in the form ____ \(\times 10^{-1}\). Let the required integer be \(Z\).

\[ Z \times 10^{-1} = \frac{t_1}{t_2} \approx 1.706 \] \[ Z = 1.706 \times 10 = 17.06 \]

The value of \(Z\) to the nearest integer is 17.

Therefore, the value of the ratio \(t_1 : t_2\) is 17 \(\times 10^{-1}\).

Answer 2

For first order reactions:

\[ K_1 t_1 = \ln\left(\frac{1}{1 - \frac{2}{3}}\right) = \ln 3 \]

\[ K_2 t_2 = \ln\left(\frac{1}{1 - \frac{4}{5}}\right) = \ln 5 \]

\[ \therefore K_1 t_1 = 5 \quad \text{and} \quad K_2 t_2 = 2 \]

\[ \frac{K_1}{K_2} = \frac{\ln 3}{\ln 5} \]

\[ \frac{t_1}{t_2} = \frac{0.477}{0.699} \times 5 = 1.7 \times 10^{-1} \]