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?

Question

Quizplus · Accepted Answer

For a second-order reaction, the rate constant $k$ can be calculated using the formula $\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt$, where $[A]_t$ is the concentration at time $t$, $[A]_0$ is the initial concentration, and $t$ is the time. Substituting the given values, $\frac{1}{3.50 \times 10^{-2}} - \frac{1}{6.75 \times 10^{-2}} = k \times 142$, solving for $k$ gives approximately $9.69 \times 10^{-2} M^{-1}s^{-1}$.

Hydrogen Iodide Decomposes at 800 K Via a Second-Order Process