Question

In a first order reaction, $\text{R} \rightarrow \text{P}$; the concentration of R changes from 1.2 M to 0.30 M in 47 min. The rate of reaction in $\text{mol L}^{-1} \text{min}^{-1}$ at a time when concentration of R is 0.2 M would be $a \times 10^{-3}$. Find the value of a.

Answer 1

6

Answer 2

The problem involves a first-order reaction, for which we need to calculate the rate constant first, and then use it to find the rate of reaction at a specific concentration.

1. Calculate the rate constant (k): For a first-order reaction, the integrated rate law is: $k = \frac{1}{t} \ln\left(\frac{[\text{R}]_0}{[\text{R}]_t}\right)$ Given: Initial concentration, $[\text{R}]_0 = 1.2$ M Concentration after time t, $[\text{R}]_t = 0.30$ M Time, $t = 47$ min

Substitute the values into the equation: $k = \frac{1}{47 \text{ min}} \ln\left(\frac{1.2 \text{ M}}{0.30 \text{ M}}\right)$ $k = \frac{1}{47} \ln(4)$

We know that $\ln(4) = 2 \ln(2)$. Using the common approximation $\ln(2) \approx 0.693$: $\ln(4) \approx 2 \times 0.693 = 1.386$

Now, calculate k: $k = \frac{1.386}{47} \text{ min}^{-1}$ $k \approx 0.029489 \text{ min}^{-1}$

2. Calculate the rate of reaction: For a first-order reaction, the rate law is: Rate $= k[\text{R}]$ We need to find the rate when the concentration of R is 0.2 M. Rate $= 0.029489 \text{ min}^{-1} \times 0.2 \text{ M}$ Rate $\approx 0.0058978 \text{ mol L}^{-1} \text{ min}^{-1}$

3. Express the rate in the required format and find 'a': The rate is given in the format $a \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}$. So, $a \times 10^{-3} = 0.0058978$ $a = \frac{0.0058978}{10^{-3}}$ $a = 5.8978$

Rounding 'a' to the nearest integer, as is common in such problems, we get $a=6$.