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: Write the zero order time.
For a zero order reaction $\text{X} \rightarrow P_1$, the integrated law is $[\text{X}] = \text{C} - k_0 t$. We want the time for X to fall to one third of C, that is to $\tfrac{\text{C}}{3}$.

Step 2: Solve for the zero order time.
Set $\tfrac{\text{C}}{3} = \text{C} - k_0 t$. Then $k_0 t = \text{C} - \tfrac{\text{C}}{3} = \tfrac{2\text{C}}{3}$, so the time is $t = \dfrac{2\text{C}}{3 k_0}$.

Step 3: Write the first order time.
For a first order reaction $\text{Y} \rightarrow P_2$, the law is $\ln\frac{[\text{Y}]_0}{[\text{Y}]} = k_1 t$. For Y falling to one third, $\frac{[\text{Y}]_0}{[\text{Y}]} = 3$.

Step 4: Solve for the first order time.
So $\ln 3 = k_1 t$, which gives $t = \dfrac{\ln 3}{k_1}$. This is the time for the first order reactant to reach a third of its start.

Step 5: Set the two times equal.
The problem says both reach one third at the same time, so we set the two expressions equal: \[ \dfrac{2\text{C}}{3 k_0} = \dfrac{\ln 3}{k_1}. \]

Step 6: Solve for the ratio $k_0/k_1$.
Cross multiply: $2\text{C}\, k_1 = 3 k_0 \ln 3$, so $\dfrac{k_0}{k_1} = \dfrac{2\text{C}}{3 \ln 3}$.

\[ \boxed{\dfrac{k_0}{k_1} = \dfrac{2\text{C}}{3\ln 3}} \]