Question
(a) Silver carbonate, $\mathrm{Ag}_2 \mathrm{CO}_3$, is sparingly soluble in water. The numerical value of the solubility product, $K_{\mathrm{sp}}$, for silver carbonate is $6.3 \times 10^{-12}$ at $25^{\circ} \mathrm{C}$.
(i) Write an expression for the solubility product, $\mathrm{K}_{\mathrm{sp}}$, of $\mathrm{Ag}_2 \mathrm{CO}_3$, and state its units.
$
K_{\mathrm{sp}}=
$ units = ………………………… [2]
(ii) Calculate the equilibrium concentration of $\mathrm{Ag}^{+}$in a saturated solution of $\mathrm{Ag}_2 \mathrm{CO}_3$ at $25^{\circ} \mathrm{C}$.
$\left[\mathrm{Ag}^{+}\right]=$ $\mathrm{moldm}^{-3}$ [1]
$\left[\mathrm{Ag}^{+}\right]=$ $\mathrm{moldm}^{-3}$ [1]
(iii) Solid $\mathrm{Ag}_2 \mathrm{CO}_3$ is stirred at $25^{\circ} \mathrm{C}$ with $0.050 \mathrm{moldm}^{-3} \mathrm{AgNO}_3$ until no more $\mathrm{Ag}_2 \mathrm{CO}_3$ dissolves. Calculate the concentration of carbonate ions, $\left[\mathrm{CO}_3{ }^{2-}\right]$, in this solution.
$
\left[\mathrm{CO}_3{ }^{2-}\right]=
$
$\mathrm{moldm}^{-3}$ [1]
(iv) An electrochemical cell is set up to measure the electrode potential, $E$, for the $\mathrm{Ag}^{+} / \mathrm{Ag}$ half-cell using the saturated $\mathrm{Ag}_2 \mathrm{CO}_3(\mathrm{aq})$ with a standard hydrogen electrode.
Use the Data Booklet, your answer to (a)(ii), and the Nernst equation to calculate the electrode potential, $E$, for this $\mathrm{Ag}^{+} / \mathrm{Ag}$ half-cell.
$E$ for $\mathrm{Ag}^{+} / \mathrm{Ag}$ half-cell $=$ $\mathrm{V}[\mathrm{s}]$
(b) Silver chloride, $\mathrm{AgCl}$, is sparingly soluble in water. The equation for the enthalpy change of solution is shown.
$
\operatorname{AgCl}(\mathrm{s}) \rightarrow \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \quad \Delta H_{\text {sol }}^{\ominus}=+65.5 \mathrm{~kJ} \mathrm{~mol}^{-1}
$
Standard entropies are shown in the table.
(i) Calculate the standard entropy change of solution, $\Delta S^\theta$.
$\Delta S^{\ominus}=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \mathrm{JK}^{-1} \mathrm{~mol}^{-1}[1]$
(ii) Explain, with the aid of a calculation, why $\mathrm{AgCl}$ is insoluble in water at $25^{\circ} \mathrm{C}$.
You should use data from this question and your answer to (b)(i).
(ii) Explain, with the aid of a calculation, why $\mathrm{AgC} l$ is insoluble in water at $25^{\circ} \mathrm{C}$. You should use data from this question and your answer to (b)(i). [3] [Total: 10]
▶️Answer/Explanation
Ans:
(a)(i) $\begin{aligned} & \text { M1 } K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CO}_3^{2-}\right] \\ & \text { M2 units }=\mathrm{mol}^3 \mathrm{dm}^{-9}\end{aligned}$
7(a)(ii) $\begin{aligned} & x=\sqrt[3]{6} .3 \times 10^{-12} / 4=1.16 \times 10^{-4}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right) \\ & {\left[\mathrm{Ag}^{+}\right]=1.16 \times 10^{-4} \times 2=2.33 \times 10^{-4}\left(\mathrm{~mol} \mathrm{dm^{-3 }}\right) \min 2 \mathrm{sf}} \\ & \end{aligned}$
(a)(iii) $
6.3 \times 10^{-12}=[0.05]^2\left[\mathrm{CO}_3{ }^{2-}\right]
$
$\left[\mathrm{CO}_3{ }^{2-}\right]=2.52 \times 10^{-9}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right) \min 2 \mathrm{sf}$
a)(iv) M1 $E=E^0+0.059 \log \left[\mathrm{Ag}^{+}\right]$
M2 $E=0.80+0.059 \log \left(1.2 \times 10^{-4}\right)=0.57 \mathrm{~V}$ ecf from (a)(ii) $\min 2 \mathrm{sf}$
b)(i) $\Delta S^{\circ}=72.7+56.5-96.2=+33.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
(b)(ii) M1 $\Delta G=\Delta H^{\ominus}-\mathrm{T} \Delta S^{\ominus}$
M2 $\Delta G=(65.5)-(298 \times 0.033)=+55.7 \mathrm{~kJ}^{-1} \mathrm{~mol}^{-1} \min 3 s f$
M3 $\Delta G=$ positive so not feasible/spontaneous
Question
One method of producing hydrogen from natural gas is the reaction between hydrogen sulfide and methane.
$
2 \mathrm{H}_2 \mathrm{~S}(\mathrm{~g})+\mathrm{CH}_4(\mathrm{~g}) \rightleftharpoons \mathrm{CS}_2(\mathrm{~g})+4 \mathrm{H}_2(\mathrm{~g})
$
(a) Write the expression for $K_p$ for this reaction, and state its units.
$
K_{\mathrm{p}}=
$
units[2]
(b) The initial partial pressures of the two gases in a mixture at $1000 \mathrm{~K}$ are recorded.
$
\mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) 200 \mathrm{~atm} \quad \mathrm{CH}_4(\mathrm{~g}) \quad 100 \mathrm{~atm}
$
The mixture is left to reach equilibrium.
It is found that the equilibrium partial pressure of $\mathrm{CS}_2(\mathrm{~g})$ is $2 \mathrm{~atm}$ and that of the remaining $\mathrm{CH}_4(\mathrm{~g})$ is $98 \mathrm{~atm}$.
(i) Calculate the equilibrium partial pressures of $\mathrm{H}_2 \mathrm{~S}(\mathrm{~g})$ and $\mathrm{H}_2(\mathrm{~g})$.
$p\left(\mathrm{H}_2 \mathrm{~S}\right)=$ atm
$p\left(\mathrm{H}_2\right)=$ atm [2]
(ii) Calculate the value of $K_{\mathrm{p}}$ at this temperature.
$
K_p=
$
(c) (i) Predict the sign of $\Delta S^{\ominus}$ for this reaction. Explain your answer.
$
2 \mathrm{H}_2 \mathrm{~S}(\mathrm{~g})+\mathrm{CH}_4(\mathrm{~g}) \rightleftharpoons \mathrm{CS}_2(\mathrm{~g})+4 \mathrm{H}_2(\mathrm{~g}) \quad \Delta H^{\ominus}=+241 \mathrm{kJmol}^{-1}
$ [1]
The free energy change, $\Delta G^{\ominus}$, for this reaction at $1000 \mathrm{~K}$ is $+51 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
(ii) Calculate the value of $\Delta S^\theta$ for this reaction, stating its units.
$\Delta S^\theta=$ units [2]
(d) How would the value of $\Delta G^{\ominus}$, and hence the spontaneity (feasibility) of this reaction change as the temperature increases? Explain your answer. [2] [Total: 10]
▶️Answer/Explanation
Ans:
(a) $\quad K_{\mathrm{p}}=\left\{\mathrm{p}\left(\mathrm{CS}_2\right) \times\left(\mathrm{p}\left(\mathrm{H}_2\right)\right)^4\right\} /\left\{\left(\mathrm{p}\left(\mathrm{H}_2 \mathrm{~S}\right)\right)^2 \times \mathrm{p}\left(\mathrm{CH}_4\right)\right\}$ units: $\mathrm{atm}^2$ OR Pa
(b) (i) $\quad \begin{aligned} & \mathrm{p}\left(\mathrm{H}_2 \mathrm{~S}\right)=196 \mathrm{~atm} \\ & \mathrm{p}\left(\mathrm{H}_2\right)=8 \mathrm{~atm}\end{aligned}$
(ii) $\quad K_p=\left(2 \times 8^4\right) /\left(196^2 \times 98\right)=2.176 \times 10^{-3}$
(c) (i) $\Delta S^{\circ}$ will be positive, because more gas moles on the RHS/products
(ii) $\quad \Delta S^{\circ}=\left(\Delta H^{\circ}-\Delta G^{\circ}\right) / T=(241-51) / 1000=0.19$ OR 190
(d) $\quad \Delta G^{\ominus}$ will become less positive/more negative as $T$ increases, …because $\Delta S^0$ is positive (or $-T \Delta S^0$ is more negative) therefore the reaction becomes more feasible/spontaneous as $T$ increases