6.2: Additional Answer Options:

2018 - May/June - Paper 2 - Question 6

6.1.1

When the equilibrium in a closed system is disturbed, the system will reinstate a new equilibrium by favouring the reaction that will oppose the disturbance.

2 marks

(2 x 1 mark)

6.1.2

Percentage yield increases with an increase in temperature.
Forward reaction is favoured.
Increase in temperature favours an endothermic reaction.

3 marks

(3 x 1 mark)

6.1.3

When the pressure increases, the reaction that leads to a decrease in the number of moles will be favoured.

When the pressure increases, the yield increases because the equilibrium position shifts to the right.

2 marks

(2 x 1 mark)

6.1.4

2 marks

(2 x 1 mark)

6.2

Option 1:

	HCl	O2	Cl2	H20
Initial quantity (mol)	0,2	0,11	0	0
Change (mol)	0,15 ⁽¹⁾	0,0375	0,075	0,075
Quantity at equilibrium (mol)	1,825/36,5 = 0,05 ⁽²⁾	0,0725	0,075	0,075
Equilibrium concentration (mol∙dm^-3)	0,25	0,3625	0,375	0,375

⁽³⁾

Ratio. ⁽⁴⁾

Divide by 0,2. ⁽⁵⁾

\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{C} \ell_2\right]^2\left[\mathrm{H}_2 \mathrm{O}\right]^2}{[\mathrm{HC} \ell]^4\left[\mathrm{O}_2\right]}=\frac{(0,375)^2(0,375)^2}{(0,25)^4(0,3625)}=13,97 ^{(6 + 7 + 8 + 9)}

There are more ways to answer this question. To view other options, click here.

9 marks

(9 x 1 mark)

Marking criteria:

⁽¹⁾Substitution of 36,5 g∙mol^-1 in n = m/M.
⁽²⁾Change n(HCℓ) = initial – equilibrium.
⁽³⁾Using ratio: 4 : 1 : 2 : 2.
⁽⁴⁾Equilibrium: n(O₂) & n(H₂O) & n(Cℓ₂) = initial ± change.
⁽⁵⁾Divide by volume (0,2 dm³).
⁽⁶⁾Correct K_cexpression (formulae in square brackets).
⁽⁷⁾Correct K_c expression (formulae in square brackets).
⁽⁸⁾Substitution of product concentrations.
⁽⁹⁾Final answer: 13,966 to 18,72.

Marking criteria for option 3:

⁽¹⁾Substitution of 36,5 g∙mol^-1 n = m/M.
⁽²⁾Initial concentration of reactants: c(HCℓ) = 1,0 & c(O2) = 0,55 mol∙dm^-3.
⁽³⁾Change: c(HCℓ) = 0,75 mol∙dm^-3 (initial – equilibrium)
⁽⁴⁾Using ratio: 4: 1: 2: 2.
⁽⁵⁾Equilibrium: c(H₂O) = c(Cℓ₂) = 0,3625 mol∙dm^-3 (initial+change) and c(O₂) = 0,3625 mol∙dm^-3 (initial – change).
⁽⁶⁾Correct Kc expression (formulae in square brackets).
⁽⁷⁾Substitution of reactant concentrations.
⁽⁸⁾Substitution of product concentrations.
⁽⁹⁾Final answer: 13,97.
Range: 13,966 to 18,72.

Option 2

$\mathrm{n}(\mathrm{HCl})_{\text{equilibrium }}=\frac{\mathrm{m}}{\mathrm{M}}=\frac{1,825}{36,5}=0,05\mathrm{mol}$ ⁽¹⁾

$\mathrm{n}(\mathrm{HCl})_{\text{reacted }}=0,2-0,05=0,15\mathrm{mol}$ ⁽²⁾

$\left.\begin{array}{l}\mathrm{n}\left(\mathrm{O}_2\right)_{\text{reacted}}=1/4\mathrm{n}(\mathrm{HCl})_{\text{reacted }}=1/4\times0,15=0,0375\mathrm{mol}\\ \mathrm{n}\left(\mathrm{Cl}_2\right)_{\text{formed}}=1/2\mathrm{n}(\mathrm{HCl})_{\text{reacted }}=1/2\times0,15=0,075\mathrm{mol}\\ \mathrm{n}\left(\mathrm{H}_2\mathrm{O}\right)_{\text{formed }}\text{ }=1/2\mathrm{n}(\mathrm{HCl})_{\text{reacted }}=1/2\times0,15=0,075\mathrm{mol}\end{array}\right\}\text{ Using ratio }$ ⁽³⁾

$\left.\begin{array}{l}\mathrm{n}\left(\mathrm{O}_2\right)_{\text{equilibrium}}=0,11-0,0375=0,0725\mathrm{mol}\\ \mathrm{n}\left(\mathrm{Cl}_2\right)_{\text{equilibrium}}=\mathrm{n}\left(\mathrm{H}_2\mathrm{O}\right)_{\text{equilibrium}}=0,075\mathrm{mol}\end{array}\right\}$ ⁽⁴⁾

$\left.\begin{array}{l}\mathrm{n}\left(\mathrm{O}_2\right)_{\text{equilibrium}}=0,11-0,0375=0,0725\mathrm{mol}\\ \mathrm{n}\left(\mathrm{Cl}_2\right)_{\text{equilibrium }}=\mathrm{n}\left(\mathrm{H}_2\mathrm{O}\right)_{\text{equilibrium}}=0,075\mathrm{mol}\end{array}\right\}\text{ Divide by 0,2}$ ⁽⁵⁾

$\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{H}_2\mathrm{O}\right]^2\left[\mathrm{C}\ell_2\right]^2}{[\mathrm{HC}\ell]^4\left[\mathrm{O}_2\right]}=\frac{(0,375)^2(0,375)^2}{(0,25)^4(0,3625)}=13,97$ ^{(6 + 7 + 8 + 9)}

Option 3

$\mathrm{n}(\mathrm{HCl})_{\text{equilibrium }}=\frac{\mathrm{m}}{\mathrm{M}}$

$=\frac{1,825}{36,5}$ ⁽¹⁾

$=0,05\mathrm{~mol}$

	HCl	O₂	H₂O	Cl₂
Initial concentration (mol∙dm^-3)	1,0 ⁽²⁾	0,55	0	0
Change in concentration (mol∙dm^-3)	0,75 ⁽³⁾	0,1875	0,375	0,375
Equilibrium concentration (mol∙dm^-3)	0,25	0,3625	0,375	0,375

Divide by 0,2. ⁽⁵⁾

Ratio. ⁽⁴⁾

$\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{C}\ell_2\right]^2\left[\mathrm{H}_2\mathrm{O}\right]^2}{[\mathrm{HC}\ell]^4\left[\mathrm{O}_2\right]}=\frac{(0,375)^2(0,375)^2}{(0,25)^4(0,3625)}=13,97$ ^{(6 + 7 + 8 + 9)}

Related subjects & topics

Grade 12 Physical Sciences

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