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 objective is to determine the ratio of \(t_1\) to \(t_2\). \(t_1\) represents the time required for \( \frac{2}{3} \) completion of a first-order reaction (Reaction 1), and \(t_2\) is the time for \( \frac{4}{5} \) completion of a different first-order reaction (Reaction 2). The ratio of their half-lives is provided.

Concept Used:

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

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

Alternatively, using base-10 logarithm:

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

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

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

This indicates that the rate constant \(k\) is inversely proportional to the half-life \(t_{1/2}\).

Step-by-Step Solution:

Step 1: Establish the relationship between the rate constants of the two reactions based on their half-life ratio.

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

Given:

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

Consequently, 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: Derive the expression for \(t_1\) for Reaction 1.

Reaction 1 proceeds to \( \frac{2}{3} \) completion, meaning \( 1 - \frac{2}{3} = \frac{1}{3} \) of the reactant remains. Thus, \( \frac{[A_t]}{[A_0]} = \frac{1}{3} \), or \( \frac{[A_0]}{[A_t]} = 3 \).

Applying 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: Derive the expression for \(t_2\) for Reaction 2.

Reaction 2 reaches \( \frac{4}{5} \) completion, so \( 1 - \frac{4}{5} = \frac{1}{5} \) of the reactant remains. Therefore, \( \frac{[P_t]}{[P_0]} = \frac{1}{5} \), 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: Compute the ratio \(t_1 : t_2\).

The ratio of \(t_1\) to \(t_2\) is:

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

From Step 1, \( \frac{k_1}{k_2} = \frac{2}{5} \), 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 provided log values to compute the final ratio.

The ratio of natural logarithms is equivalent 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 requires the ratio to be expressed in the format \( Z \times 10^{-1} \). Let the required integer be \(Z\).

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

Rounding \(Z\) to the nearest integer yields 17.

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