Chemical equilibrium — Question 2

2a)

Exothermic Increasing temperature decreases K_c, indicating that the equilibrium shifts towards the reactants (endothermic) direction.

• determines forward reaction is exothermic [1 mark]

• explain that the decrease in K_c as temperature increases indicates endothermic direction is towards the reactants [1 mark]

2b)

$_{11}[\mathrm{S}\mathrm{O}_{3}]^{2}$

$8.61\,\times 10=_{2}\,$

$(0.860)\,(0.330)$

$[\mathrm{S}\mathrm{O}]^{2}=(0.7396)(0.330)8.61\times 10^{11}=2.10\times 10^{11}\,$

$_{3}()$

$\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\,=\,4.58\,\times \,10^{5}\,\mathrm{M}\,$

$•\,\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{v}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{s}\,\mathrm{a}\mathrm{p}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{e}\,\mathrm{w}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{g}\,[1\,\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}]\,$

$•\,\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{s}\,[\mathrm{S}\mathrm{O}_{3}]\,=\,4.58\,\times \,10^{5}\,[1\,\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}]\,$

2c)

Halving the volume would double the pressure. To reduce the pressure, the equilibrium would shift toward the product to reduce the number of molecules present. However, due to the reactant decreasing the equilibrium constant remains unchanged.

• indicates that halving the volume doubles the pressure [1 mark]

• explains that equilibrium will shift to reduce the number of molecules present to reduce pressure [1 mark]

• explains that equilibrium will shift toward the product [1 mark]

• explains why the equilibrium constant would not change [1 mark]