| 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.
\(\begin{array}{*{20}{l}} {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{4}}}{{{\text{(OH)}}}_{\text{2}}}{\text{(aq)}} \to {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{4}}}{{\text{O}}_{\text{2}}}{\text{(aq)}} + {{\text{H}}_{\text{2}}}{\text{(g)}}}&{\Delta {H^\theta } = + {\text{177.0 kJ}}} \\ {{\text{2}}{{\text{H}}_{\text{2}}}{\text{O(l)}} + {{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{(aq)}}}&{\Delta {H^\theta } = + {\text{189.2 kJ}}} \\ {{{\text{H}}_{\text{2}}}{\text{O(l)}} \to {{\text{H}}_{\text{2}}}{\text{(g)}} + \frac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}{\text{(g)}}}&{\Delta {H^\theta } = + {\text{285.5 kJ}}} \end{array}\)
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.)
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 – \(\frac{{189.2}}{2}\) –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]
203.1 «kJ» = 0.850 «kg» x 4.18 «kJ\(\,\)kg–1\(\,\)K–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 «kJ\(\,\)kg–1\(\,\)K–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]
