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Date	May 2017	Marks available	1	Reference code	17M.2.sl.TZ2.8
Level	SL	Paper	2	Time zone	TZ2
Command term	Identify	Question number	8	Adapted from	N/A

Question

The Bombardier beetle sprays a mixture of hydroquinone and hydrogen peroxide to fight off predators. The reaction equation to produce the spray can be written as:

C₆H₄(OH)₂(aq) + H₂O₂(aq)	→	C₆H₄O₂(aq) + 2H₂O(l)
hydroquinone		quinone

Calculate the enthalpy change, in kJ, for the spray reaction, using the data below.

$\begin{array}{*{20}{l}} {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{4}}}{{{\text{(OH)}}}_{\text{2}}}{\text{(aq)}} \to {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{4}}}{{\text{O}}_{\text{2}}}{\text{(aq)}} + {{\text{H}}_{\text{2}}}{\text{(g)}}}&{\Delta {H^\theta } = + {\text{177.0 kJ}}} \\ {{\text{2}}{{\text{H}}_{\text{2}}}{\text{O(l)}} + {{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{(aq)}}}&{\Delta {H^\theta } = + {\text{189.2 kJ}}} \\ {{{\text{H}}_{\text{2}}}{\text{O(l)}} \to {{\text{H}}_{\text{2}}}{\text{(g)}} + \frac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}{\text{(g)}}}&{\Delta {H^\theta } = + {\text{285.5 kJ}}} \end{array}$

[2]

a.i.

The energy released by the reaction of one mole of hydrogen peroxide with hydroquinone is used to heat 850 cm³ of water initially at 21.8°C. Determine the highest temperature reached by the water.

Specific heat capacity of water = 4.18 kJ $\,$ kg⁻¹ $\,$ K⁻¹.

(If you did not obtain an answer to part (i), use a value of 200.0 kJ for the energy released, although this is not the correct answer.)

[2]

a.ii.

Identify the species responsible for the peak at m/z = 110 in the mass spectrum of hydroquinone.

[1]

Identify the highest m/z value in the mass spectrum of quinone.

[1]

Markscheme

ΔH = 177.0 – $\frac{{189.2}}{2}$ –285.5 «kJ»

«ΔH =» –203.1 «kJ»

Accept other methods for correct manipulation of the three equations.

Award [2] for correct final answer.

[2 marks]

a.i.

203.1 «kJ» = 0.850 «kg» x 4.18 «kJ $\,$ kg^{–1 $\,$}K^–1» x ΔT «K»
OR
«ΔT =» 57.2 «K»
«T_final = (57.2 + 21.8) °C =» 79.0 «°C» / 352.0 «K»

If 200.0 kJ was used:
200.0 «kJ» = 0.850 «kg» x 4.18 «kJ $\,$ kg^–1 $\,$ K^–1» x ΔT «K»
OR
«ΔT =» 56.3 «K»
«T_final = (56.3 + 21.8) °C =» 78.1 «°C» / 351.1 «K»