| Date | November 2009 | Marks available | 4 | Reference code | 09N.2.sl.TZ0.3 |
| Level | SL | Paper | 2 | Time zone | TZ0 |
| Command term | Calculate | Question number | 3 | Adapted from | N/A |
Question
The standard enthalpy change of three combustion reactions is given below in kJ.
\[\begin{array}{*{20}{l}} {{\text{2}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{(g)}} + {\text{7}}{{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{4C}}{{\text{O}}_{\text{2}}}{\text{(g)}} + {\text{6}}{{\text{H}}_{\text{2}}}{\text{O(l)}}}&{\Delta {H^\Theta } = - 3120} \\ {{\text{2}}{{\text{H}}_2}({\text{g)}} + {{\text{O}}_2}{\text{(g)}} \to {\text{2}}{{\text{H}}_2}{\text{O(l)}}}&{\Delta {H^\Theta } = - 572} \\ {{{\text{C}}_2}{{\text{H}}_4}({\text{g)}} + {\text{3}}{{\text{O}}_2}{\text{(g)}} \to {\text{2C}}{{\text{O}}_2}{\text{(g)}} + 2{{\text{H}}_2}{\text{O(l)}}}&{\Delta {H^\Theta } = - 1411} \end{array}\]
Based on the above information, calculate the standard change in enthalpy, \({\Delta {H^\Theta }}\), for the following reaction.
\[{{\text{C}}_2}{{\text{H}}_6}({\text{g)}} \to {{\text{C}}_2}{{\text{H}}_4}({\text{g)}} + {{\text{H}}_2}{\text{(g)}}\]
Markscheme
\(\begin{array}{*{20}{l}} {\left( {{{\text{C}}_2}{{\text{H}}_6}{\text{(g)}} + {\text{3}}\frac{1}{2}{{\text{O}}_2}{\text{(g)}} \to {\text{2C}}{{\text{O}}_2}{\text{(g)}} + {\text{3}}{{\text{H}}_2}{\text{O(l)}}} \right)}&{\Delta {H^\Theta } = - 1560;} \\ {\left( {{{\text{H}}_2}{\text{O(l)}} \to {{\text{H}}_2}{\text{(g)}} + \frac{1}{2}{{\text{O}}_2}{\text{(g)}}} \right)}&{\Delta {H^\Theta } = + 286;} \\ {\left( {{\text{2C}}{{\text{O}}_2}{\text{(g)}} + {\text{2}}{{\text{H}}_2}{\text{O(l)}} \to {{\text{C}}_2}{{\text{H}}_4}{\text{(g)}} + {\text{3}}{{\text{O}}_2}{\text{(g)}}} \right)}&{\Delta {H^\Theta } = + 1411;} \\ {\left( {{{\text{C}}_2}{{\text{H}}_6}{\text{(g)}} \to {{\text{C}}_2}{{\text{H}}_4}{\text{(g)}} + {{\text{H}}_2}{\text{(g)}}} \right)}&{\Delta {H^\Theta } = + 137{\text{ (kJ)}};} \end{array}\)
Allow other correct methods.
Award [2] for –137.
Allow ECF for the final marking point.
Examiners report
A significant number of candidates scored full marks here. Although the setting out of the calculations was often difficult to follow, a pleasing number of correct final answers was seen. The commonest errors were to give a correct numerical value with a negative sign and to fail to divide the value for the first equation by 2.
