In the following first order competing reactionsthe ratio of  k1k2 , if only  94%  of  A  have been reacted whereas  50%  of  C has been reacted, is

To determine the ratio $\frac{k_1}{k_2}$ for the given first-order competing reactions, we need to analyze the reaction progress based on the information provided.

1. ### Break Down the Problem

• We have two competing reactions, where A converts into B (with rate constant $k_1$) and C (with rate constant $k_2$).
• Given that 94% of A has reacted, this means 6% of A remains.
• Given that 50% of C has reacted.

2. ### Relevant Concepts

For first-order reactions, the relationship between the amount reacted and the rate constants can be described using the following formula for the fraction remaining:

$\frac{[A]_t}{[A]_0} = e^{-k_1 t} \quad \text{for A}$

$\frac{[C]_t}{[C]_0} = e^{-k_2 t} \quad \text{for C}$

Where:

• $[A]_t$ is the concentration of A at time $t$,
• $[A]_0$ is the initial concentration of A,
• $[C]_t$ and $[C]_0$ are defined similarly for C.

3. ### Analysis and Detail

1. From the information:

   • 94% of A has reacted, therefore 6% remains: $\frac{[A]_t}{[A]_0} = 0.06 \implies e^{-k_1 t} = 0.06$ Taking the natural logarithm: $-k_1 t = \ln(0.06) \implies k_1 t = -\ln(0.06)$

2. For C, 50% has reacted, meaning 50% remains: $\frac{[C]_t}{[C]_0} = 0.50 \implies e^{-k_2 t} = 0.50$ Again, taking the natural logarithm: $-k_2 t = \ln(0.50) \implies k_2 t = -\ln(0.50)$

4. ### Verify and Summarize

Now we have the two equations:

1. $k_1 t = -\ln(0.06)$
2. $k_2 t = -\ln(0.50)$

To find the ratio $\frac{k_1}{k_2}$: $\frac{k_1}{k_2} = \frac{-\ln(0.06)}{-\ln(0.50)}$

Calculating the logarithms: $\ln(0.06) \approx -2.813 \quad \text{and} \quad \ln(0.50) \approx -0.693$

Therefore: $\frac{k_1}{k_2} = \frac{2.813}{0.693} \approx 4.06$

Final Answer

The ratio $\frac{k_1}{k_2}$ is approximately $4.06$.