Question

In a zero order reaction C→P, 37.5% of the reactant remains at the end of 2.5 hours. The percentage of reactant consumed in 12 minutes is:

Answer 1

5

Answer 2

The problem describes a zero-order reaction. For a zero-order reaction, the rate of consumption of the reactant is constant, meaning the amount of reactant consumed is directly proportional to time.

The integrated rate law for a zero-order reaction can be expressed as: $[A]_t = [A]_0 - kt$ where: $[A]_t$ is the concentration of reactant at time t $[A]_0$ is the initial concentration of reactant $k$ is the rate constant $t$ is the time

Alternatively, the amount of reactant consumed ($x$) at time $t$ is given by: $x = kt$

Step 1: Calculate the rate constant (k) Given: Initial percentage of reactant, $[A]_0 = 100\%$ Percentage of reactant remaining at $t_1 = 2.5$ hours, $[A]_{t_1} = 37.5\%$

The percentage of reactant consumed ($x_1$) in 2.5 hours is: $x_1 = [A]_0 - [A]_{t_1}$ $x_1 = 100\% - 37.5\% = 62.5\%$

Now, use the formula $x = kt$ to find $k$: $62.5\% = k \times 2.5 \text{ hours}$ $k = \frac{62.5\%}{2.5 \text{ hours}}$ $k = 25\%/\text{hour}$

Step 2: Calculate the percentage of reactant consumed in 12 minutes First, convert 12 minutes to hours to match the unit of $k$: $t_2 = 12 \text{ minutes} = \frac{12}{60} \text{ hours} = 0.2 \text{ hours}$

Now, use the formula $x = kt$ to find the percentage of reactant consumed ($x_2$) in 12 minutes: $x_2 = k \times t_2$ $x_2 = (25\%/\text{hour}) \times (0.2 \text{ hours})$ $x_2 = 25 \times \frac{1}{5} \%$ $x_2 = 5\%$

The percentage of reactant consumed in 12 minutes is 5%.

The final answer is $\boxed{5}$.

Explanation: For a zero-order reaction, the rate of consumption is constant. We first determine this constant rate (k) using the given information: 62.5% of the reactant is consumed in 2.5 hours, which implies a rate of 25% per hour. Then, we use this rate to find the amount consumed in the desired time of 12 minutes (0.2 hours), resulting in 5%.