Explain what each of the following terms means: homogeneous catalyst heterogeneous catalyst
Iodide ions react with peroxydisulfate ions. $\text{2I}^- + \text{S}_2\text{O}_8^{2-} \rightarrow \text{I}_2 + 2\text{SO}_4^{2-}$ The uncatalysed reaction is slow, but $\text{Fe}^{2+}$ ions catalyse it. Write two equations that show how $\text{Fe}^{2+}$ ions catalyse this reaction.
Suggest why the alternative pathway in the presence of $\text{Fe}^{2+}$ ions has a smaller activation energy than the pathway without a catalyst.
Nitrogen monoxide reacts with oxygen. $2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)$ This reaction is second order in nitrogen monoxide and first order in oxygen. Write the rate equation for this reaction.
Calculate the initial rate of reaction under these conditions when the starting concentration of nitrogen monoxide is $7.20 \times 10^{-4}\,\text{mol dm}^{-3}$ and the starting concentration of oxygen is $1.90 \times 10^{-3}\,\text{mol dm}^{-3}$. For the stated conditions, the rate constant $k$ is $8.60 \times 10^6\,\text{mol}^{-2}\,\text{dm}^6\,\text{s}^{-1}$.
The drug cisplatin, $\text{Pt(NH}_3)_2\text{Cl}_2$, hydrolyses in water. $\text{Pt(NH}_3)_2\text{Cl}_2 + \text{H}_2\text{O} \rightarrow [\text{Pt(NH}_3)_2\text{Cl(H}_2\text{O})]^+ + \text{Cl}^-$ The rate law is given below. $\text{rate} = k[\text{Pt(NH}_3)_2\text{Cl}_2]$ Under certain conditions, $k = 2.50 \times 10^{-5}\ \text{s}^{-1}$. This reaction has a constant half-life. Explain why this happens.
Use the information in this question to show that the half-life for this reaction is $2.77 \times 10^4\ \text{s}$.
$8.00 \times 10^{-6}$ moles of $\text{Pt(NH}_3)_2\text{Cl}_2$ are added to $100\ \text{cm}^3$ of water. Calculate the time needed for the concentration of $\text{Pt(NH}_3)_2\text{Cl}_2$ to drop to $2.50 \times 10^{-6}\ \text{mol dm}^{-3}$.