Question

Consider two reactions.
I. A zero order reaction: $\text{X} \rightarrow P_1$.
II. A first order reaction: $\text{Y} \rightarrow P_2$.
Both reactions begin simultaneously with the same initial concentration `C', that is $\text{X}_0 = \text{Y}_0 = \text{C}$. It is observed that the concentrations of the reactants fall to one-third of their initial values at the same time, provided the rate constants satisfy a certain ratio. If $k_0$ and $k_1$ denote, respectively, the zero and first order rate constants then the ratio $k_0/k_1$ necessary for this condition to be satisfied is:

• A. $\frac{2\text{C}}{3\ln 3}$
• B. $\frac{\text{C}}{\ln 3}$
• C. $\frac{\text{C}}{\ln 2}$
• D. $\frac{3\text{C}}{2\ln 2}$

Hint:

Always write out the explicit expressions for time ($t$) for each kinetic order first, then equate them. This avoids algebraic errors when dealing with logarithmic terms.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the exact ratio of the rate constants ($k_0/k_1$) of a zero-order and a first-order reaction such that both starting reactants, with the same initial concentration C, fall to one-third of their initial concentration at the exact same time ($t$).

Step 2: Key Formula or Approach:
1. Integrated rate law for a zero-order reaction:
\[ [\text{X}]_t = [\text{X}]_0 - k_0 t \]
2. Integrated rate law for a first-order reaction:
\[ \ln\left(\frac{[\text{Y}]_0}{[\text{Y}]_t}\right) = k_1 t \implies t = \frac{1}{k_1} \ln\left(\frac{[\text{Y}]_0}{[\text{Y}]_t}\right) \]

Step 3: Detailed Explanation:

• Let us analyze Reaction I (zero-order): $\text{X} \rightarrow P_1$ with initial concentration $[\text{X}]_0 = \text{C}$.

• The concentration falls to one-third of its initial value, so:
\[ [\text{X}]_t = \frac{\text{C}}{3} \]

• Substituting this into the zero-order integrated rate law:
\[ \frac{\text{C}}{3} = \text{C} - k_0 t \]
\[ k_0 t = \text{C} - \frac{\text{C}}{3} = \frac{2\text{C}}{3} \]

• Solving for the time ($t$):
\[ t = \frac{2\text{C}}{3 k_0} \]

• Let us analyze Reaction II (first-order): $\text{Y} \rightarrow P_2$ with initial concentration $[\text{Y}]_0 = \text{C}$.

• The concentration falls to one-third of its initial value, so:
\[ [\text{Y}]_t = \frac{\text{C}}{3} \]

• Substituting this into the first-order integrated rate law:
\[ t = \frac{1}{k_1} \ln\left(\frac{\text{C}}{\text{C}/3}\right) = \frac{\ln 3}{k_1} \]

• Since both reactions fall to one-third of their initial values at the exact same time ($t$), we set the two expressions for $t$ equal to each other:
\[ \frac{2\text{C}}{3 k_0} = \frac{\ln 3}{k_1} \]

• Rearranging this equation to solve for the ratio $\frac{k_0}{k_1}$:
\[ \frac{k_1}{k_0} = \frac{3 \ln 3}{2\text{C}} \implies \frac{k_0}{k_1} = \frac{2\text{C}}{3\ln 3} \]


Step 4: Final Answer:
Therefore, the ratio of the rate constants $k_0/k_1$ is $\frac{2\text{C}}{3\ln 3}$ (Option A).