Date | May 2017 | Marks available | 2 | Reference code | 17M.2.sl.TZ2.8 |
Level | SL | Paper | 2 | Time zone | TZ2 |
Command term | Determine | 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.
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 kJkg−1K−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.)
Identify the species responsible for the peak at m/z = 110 in the mass spectrum of hydroquinone.
Identify the highest m/z value in the mass spectrum of quinone.
Markscheme
ΔH = 177.0 – –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]
203.1 «kJ» = 0.850 «kg» x 4.18 «kJkg–1K–1» x ΔT «K»
OR
«ΔT =» 57.2 «K»
«Tfinal = (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 «kJkg–1K–1» x ΔT «K»
OR
«ΔT =» 56.3 «K»
«Tfinal = (56.3 + 21.8) °C =» 78.1 «°C» / 351.1 «K»
Award [2] for correct final answer.
Units, if specified, must be consistent with the value stated.
[2 marks]
C6H4(OH)2+
Accept “molecular ion”.
Do not accept “C6H4(OH)2” (positive charge missing).
[1 mark]
«highest m/z» 108
Only accept exactly 108, not values close to this.
[1 mark]