The reaction: 2 HI → H2 + I2, is second order and the rate constant at 800 K is 9.70 × 10-2 M-1 s-1. How long will it take for 8.00 × 10-2 mol/L of HI to decrease to one-fourth of its initial concentration?
A) 0.619 s
B) 124 s
C) 387 s
D) 429 s
The correct answer and explanation is :
The given reaction is:
$$
2 HI \rightarrow H_2 + I_2
$$
We are told that the reaction is second order with respect to HI, and the rate constant at 800 K is $k = 9.70 \times 10^{-2} \, \text{M}^{-1} \text{s}^{-1}$. The initial concentration of HI is $[HI]_0 = 8.00 \times 10^{-2} \, \text{mol/L}$, and we want to determine the time it will take for the concentration of HI to decrease to one-fourth of its initial value.
Key Concept:
For a second-order reaction, the integrated rate law is given by:
$$
\frac{1}{[A]} – \frac{1}{[A]_0} = kt
$$
Where:
- $[A]$ is the concentration of the reactant (HI) at time $t$,
- $[A]_0$ is the initial concentration of the reactant,
- $k$ is the rate constant,
- $t$ is the time.
Step 1: Set up the equation
We are given that the concentration of HI decreases to one-fourth of its initial concentration. Therefore, at the time we are solving for, the concentration $[HI] = \frac{1}{4}[HI]_0$. Substituting the known values:
$$
\frac{1}{[HI]} – \frac{1}{[HI]_0} = kt
$$
Substitute $[HI] = \frac{1}{4}[HI]_0$, so:
$$
\frac{1}{\frac{1}{4}[HI]_0} – \frac{1}{[HI]_0} = kt
$$
Simplifying:
$$
\frac{4}{[HI]_0} – \frac{1}{[HI]_0} = kt
$$
$$
\frac{3}{[HI]_0} = kt
$$
Step 2: Solve for $t$
Rearranging the equation:
$$
t = \frac{3}{k [HI]_0}
$$
Substitute the given values:
$$
t = \frac{3}{(9.70 \times 10^{-2} \, \text{M}^{-1} \text{s}^{-1})(8.00 \times 10^{-2} \, \text{mol/L})}
$$
$$
t = \frac{3}{(7.76 \times 10^{-3} \, \text{mol}^{-1} \text{L} \text{s})}
$$
$$
t \approx 387 \, \text{s}
$$
Final Answer:
The correct answer is C) 387 s.
Explanation:
For a second-order reaction, the rate law depends on the inverse of the concentration. By using the integrated rate law for second-order reactions, we can determine how long it takes for the concentration of HI to decrease to one-fourth of its initial concentration. Through substitution and algebraic manipulation, we find that the time required is approximately 387 seconds. This demonstrates the relationship between the concentration of a reactant and the time needed for a specific change in concentration in second-order kinetics.