The second-order decomposition of HI has a rate constant of 1.80 × 10^-3 L mol^-1s^-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 mol L^-1?

Question

Quizplus · Accepted Answer

For a second-order reaction, the integrated rate law is 1/[A] = kt + 1/[A]₀. Plugging in the values: 1/[A] = (1.80 × 10⁻³ L mol⁻¹ s⁻¹)(27.3 s) + 1/4.78 mol L⁻¹. Solving for [A] gives approximately 3.87 mol L⁻¹.

The Second-Order Decomposition of HI Has a Rate Constant of 1.80