A small quantity of $\text{H}_2(g)$ is combined with a large excess of $\text{I}_2(g)$ at $400\,\text{K}$, and the reaction is followed. The graph produced is shown.
Suggest why a large excess of $\text{I}_2(g)$ is chosen for this experiment.
The reaction is first order with respect to $\text{H}_2(g)$. Use graph data to support this claim.
At $400\,\text{K}$, three separate experiments were done using different starting concentrations of $\text{H}_2(g)$ and $\text{I}_2(g)$. The results are given in the table.
Use the data, together with the order of reaction with respect to $\text{H}_2(g)$ stated in (a)(ii), to deduce the order of reaction with respect to $\text{I}_2(g)$. Explain your reasoning, and support it with data.
Use information from (a)(ii) together with your answer to (b)(i) to write the rate equation for the forward reaction.
Use your rate equation together with the data from experiment $1$ to calculate the value of the rate constant, $k$, for the forward reaction at $400\,\text{K}$. Include the units for $k$.
At $400\,\text{K}$, the forward reaction rate constant is about $1000$ times larger than the backward reaction rate constant. The total orders of the forward and backward reactions are identical. Forward reaction: $\text{H}_2(g) + \text{I}_2(g) \rightarrow \text{2HI}(g)$ Backward reaction: $\text{2HI}(g) \rightarrow \text{H}_2(g) + \text{I}_2(g)$
Use this information to explain what happens if equal concentrations of $\text{HI}(g)$, $\text{H}_2(g)$ and $\text{I}_2(g)$ are mixed at $400\,\text{K}$. You should comment on: - the relative initial rates of the forward and backward reactions - where the equilibrium lies.
At $700\,\text{K}$, the rate constant for the forward reaction is approximately $50$ times greater than the rate constant for the backward reaction. Use this information together with the information in (c)(i) to deduce the signs of the $\Delta H$ values for the forward and backward reactions. Explain your answer.