Define enthalpy change of solution, $\Delta H_{\text{sol}}$.
KI(s) has a high solubility in water although its enthalpy change of solution is endothermic. Explain how this high solubility is possible.
Table 1.1 gives some data about the halide ions, $\text{Cl}^-$, $\text{Br}^-$ and $\text{I}^-$, and their potassium salts.
Explain the trend in the enthalpy change of hydration of the halide ions.
The $\Delta H_{\text{sol}}$ values of these potassium halides are almost constant. Use the $\Delta H_{\text{hyd}}$ and $\Delta H_{\text{latt}}$ data in Table 1.1 to suggest why.
The enthalpy change of solution of KI(s) is $+21.0\,\text{kJ mol}^{-1}$. Use this information and the data in Table 1.1 to calculate the enthalpy change of hydration of the potassium ion, $\text{K}^+(g)$. Give your answer for $\Delta H_{\text{hyd}}$ of $\text{K}^+(g)$ in $\text{kJ mol}^{-1}$.
Solid $\text{PbI}_2$ forms when KI(aq) is mixed with $\text{Pb}^{2+}$(aq) ions. The solubility product, $K_{sp}$, of $\text{PbI}_2$ is $7.1 \times 10^{-9}\,\text{mol}^3\,\text{dm}^{-9}$ at $25^{\circ}\text{C}$. Calculate the solubility, in $\text{mol dm}^{-3}$, of $\text{PbI}_2$(s).
The ionic radius of $\text{Pb}^{2+}$ is $0.120\,\text{nm}$ compared to $0.133\,\text{nm}$ for $\text{K}^+$. Suggest how the $\Delta H^{\circ}_{\text{latt}}$ of $\text{PbI}_2$(s) differs from $\Delta H^{\circ}_{\text{latt}}$ of KI(s). Explain your answer.
KI slowly oxidises in air, producing $\text{I}_2$. Reaction 1 is $4\text{KI}(s) + 2\text{CO}_2(g) + \text{O}_2(g) \rightarrow 2\text{K}_2\text{CO}_3(s) + 2\text{I}_2(s)$ with $\Delta H^{\circ} = -203.4\,\text{kJ mol}^{-1}$. Table 1.2 contains some data relevant to this question.
Calculate the standard entropy change, $\Delta S^{\circ}$, of reaction 1.
Use your answer to c(i) to show that reaction 1 is spontaneous at $298\,\text{K}$.
The Group 1 carbonates are much more thermally stable than the Group 2 carbonates. State and explain the trend in the thermal stability of the Group 2 carbonates.
A student electrolyses a solution of $\text{KI}(aq)$ for 8 minutes using a direct current. The half‑equation at the anode is $2\text{I}^-(aq) \rightarrow \text{I}_2(aq) + 2e^-$. Write a half‑equation for the reaction that occurs at the cathode. Include state symbols.
After the electrolysis, the $\text{I}_2(aq)$ produced requires $21.35\,\text{cm}^3$ of $0.100\,\text{mol dm}^{-3}\,\text{Na}_2\text{S}_2\text{O}_3(aq)$ to react completely. The reaction is $\text{I}_2(aq) + 2\text{Na}_2\text{S}_2\text{O}_3(aq) \rightarrow 2\text{NaI}(aq) + \text{Na}_2\text{S}_4\text{O}_6(aq)$. Calculate the average current used in 8 minutes during the electrolysis.
KI is used as a source of $\text{I}^-$ ions in organic synthesis. One example is shown in Fig. 1.1. Identify the reagents required for steps 1 and 2.
Step 3 occurs in two stages. stage I $\text{NaNO}_2$ and $\text{HCl}$ first undergo an acid-base reaction to make $\text{HNO}_2$. stage II $\text{HNO}_2$ then reacts with $\text{C}$, $\text{C}_6\text{H}_5\text{NH}_2$, to produce $\text{D}$, $\text{C}_6\text{H}_5\text{N}_2^+$. Complete the equations for stage I and for stage II. stage I: $\text{NaNO}_2 + \text{HCl} \rightarrow \text{........}$ stage II: $\text{........}$
The $\text{I}^-$ from KI reacts with $\text{D}$ in step 4. The mechanism is shown in Fig. 1.1. Suggest the name for this mechanism.
Step 3 takes place in two stages. stage I $\text{NaNO}_2$ and $\text{HCl}$ first undergo an acid-base reaction to make $\text{HNO}_2$. stage II $\text{HNO}_2$ then reacts with $\text{C}$, $\text{C}_6\text{H}_5\text{NH}_2$, to produce $\text{D}$, $\text{C}_6\text{H}_5\text{N}_2^+$. Complete the equations for stage I and for stage II. stage I: $\text{NaNO}_2 + \text{HCl} \rightarrow \text{........}$ stage II: $\text{........}$
The $\text{I}^-$ from $\text{KI}$ reacts with $\text{D}$ in step 4. The mechanism is shown in Fig. 1.1. Suggest the name for this mechanism.