Chapter 11: Problem 99
The gas-phase reaction between hydrogen and iodine $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$$ proceeds with a rate constant for the forward reaction at \(700^{\circ} \mathrm{C}\) of \(138 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) and an activation energy of \(165 \mathrm{~kJ} / \mathrm{mol}\). (a) Calculate the activation energy of the reverse reaction given that \(\Delta H_{i}^{\circ}\) for HI is \(26.48 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta H_{i}^{\circ}\) for \(\mathrm{I}_{2}(\mathrm{~g})\) is \(62.44 \mathrm{~kJ} / \mathrm{mol}\). (b) Calculate the rate constant for the reverse reaction at \(700^{\circ} \mathrm{C}\). (Assume \(\mathrm{A}\) in the equation \(k=\mathrm{Ae}^{-\mathcal{P}_{2} / \mathrm{RT}^{\prime}}\) is the same for both forward and reverse reactions.) (c) Calculate the rate of the reverse reaction if the concentration of HI is \(0.200 M\). The reverse reaction is second-order in HI.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.