IB Chemistry HL 2.3 Extent of Chemical Change Question Bank

«moles of» ethanol «at equilibrium» =0.2 «moles of» ethyl ethanoate «at equilibrium» =0.4 «moles of» water «at equilibrium» =0.4

$\frac{0.4^{2}}{0.2^{2}}$$$\frac{<0.16 \gg}{<0.04>} 4.0$$

Marking guidance:

Award [3] for correct final answer.

M1 can be awarded for a correct expression for M2.

«the value of K in (e)(i) is too low and» insufficient time was given for equilibrium to be
reached.

reaction quotient / Q

rates of forward and reverse reactions are equal / opposing changes occur at equal rates; the concentrations of all reactants and products remain constant / macroscopic properties remain constant; closed/isolated system;

Marking guidance:

Accept "the same" for "equal" in M1 and for "constant" in M2.

The volume of the container is increased: position of equilibrium shifts to the left/reactants and fewer moles of gas on the right hand side/pressure decreases / OWTTE;

Ammonia is removed from the equilibrium mixture: position of equilibrium shifts to the right/products and [ $\mathrm{NH}_{3}$ ] decreases so [ $\mathrm{N}_{2}$ ] and [ $\mathrm{H}_{2}$ ] must also decrease to keep $K_{\mathrm{c}}$ constant OR
position of equilibrium shifts to the right/products and rate of reverse reaction decreases / OWTTE;

Marking guidance:

Award [1 max] if both predicted changes are correct.
Do not accept "to increase [ $\mathrm{NH}_{3}$ ]" or reference to LCP without explanation.

$\left(K_{\mathrm{c}}=\right) \frac{\left[\mathrm{NH}_{3}(\mathrm{~g})\right]^{2}}{\left[\mathrm{~N}_{2}(\mathrm{~g})\right] \times\left[\mathrm{H}_{2}(\mathrm{~g})\right]^{3}}$;

Marking guidance:

Ignore state symbols.
Concentrations must be represented by square brackets.

moles at equilibrium: nitrogen 0.27 , hydrogen 0.81 / concentrations at equilibrium: nitrogen $0.27\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)$, hydrogen $0.81\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)$ (and ammonia $1.46 \mathrm{~mol} \mathrm{dm}^{-3}$ );
$K_{\mathrm{c}}=15$;

Actual calculation gives $K_{c}=14.86$.

Marking guidance:

Award [2] for correct final answer.
Award [1 max] if $K_{\mathrm{c}}\left(=\frac{1.46^{2}}{3^{3} \times 1}\right)=0.079$

$\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{Br}_{2}(\mathrm{l})$;
$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})$;

increase volume of liquid / no change of colour of vapour;

shifts to right/toward products/forward reaction favoured;

Marking guidance:

Accept reverse statement if process written the other way around. Answer must match stated equation.

no effect; same amounts/number of (gaseous) moles/molecules on both sides;

$\left(K_{\mathrm{c}}=\right) \frac{[\mathrm{HBr}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]}$;

no effect (only depends on the temperature);