Hydrogen iodide decomposes at 800 K via a second-order process to produce hydrogen and iodine according to the following chemical equation.
2 HI(g) → H2(g) + I2(g)
At 800 K it takes 142 seconds for the initial concentration of HI to decrease from 6.75 × 10-2 M to 3.50 × 10-2 M. What is the rate constant for the reaction at this temperature?
A) 5.12 × 10-4 M-1s-1
B) 9.69 × 10-2 M-1s-1
C) 10.3 M-1s-1
D) 1.95 × 103 M-1s-1
The correct answer and explanation is :
To solve this problem, we need to use the integrated rate law for a second-order reaction. For a reaction of the form:
$$
2 \text{HI}(g) \rightarrow \text{H}_2(g) + \text{I}_2(g)
$$
where the reaction is second-order with respect to HI, the integrated rate law is given by:
$$
\frac{1}{[A]} – \frac{1}{[A_0]} = kt
$$
Where:
- $[A]$ is the concentration of HI at time $t$,
- $[A_0]$ is the initial concentration of HI,
- $k$ is the rate constant,
- $t$ is the time.
Step 1: Given information
From the problem, we are given the following data:
- Initial concentration of HI: $[A_0] = 6.75 \times 10^{-2} \, \text{M}$
- Concentration of HI after 142 seconds: $[A] = 3.50 \times 10^{-2} \, \text{M}$
- Time: $t = 142 \, \text{seconds}$
We are asked to find the rate constant $k$ for this reaction at 800 K.
Step 2: Apply the second-order rate law
Substitute the known values into the integrated rate law:
$$
\frac{1}{[A]} – \frac{1}{[A_0]} = kt
$$
Substituting the values:
$$
\frac{1}{3.50 \times 10^{-2}} – \frac{1}{6.75 \times 10^{-2}} = k \times 142
$$
Now, calculate the values:
$$
\frac{1}{3.50 \times 10^{-2}} = 28.57 \, \text{M}^{-1}
$$
$$
\frac{1}{6.75 \times 10^{-2}} = 14.81 \, \text{M}^{-1}
$$
$$
28.57 – 14.81 = k \times 142
$$
$$
13.76 = k \times 142
$$
Solve for $k$:
$$
k = \frac{13.76}{142} = 9.69 \times 10^{-2} \, \text{M}^{-1}\text{s}^{-1}
$$
Step 3: Conclusion
The rate constant $k$ for the reaction is $9.69 \times 10^{-2} \, \text{M}^{-1}\text{s}^{-1}$. Therefore, the correct answer is:
B) 9.69 × 10^-2 M^-1s^-1
Explanation:
This problem involves using the second-order integrated rate law to find the rate constant. The second-order reaction’s integrated rate law expresses the inverse of the concentration of the reactant at any given time. By substituting the known concentrations and time into this equation, we can solve for the rate constant. This method allows us to calculate the rate constant for reactions when the concentration changes over time, which is useful for understanding how quickly a reaction occurs under specific conditions.