Consider the equations:

$$\begin{array}{*{20}{l}} {{{\text{N}}_{\text{2}}}{\text{(g)}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{(g)}} \to {{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}{\text{(l)}}}&{\Delta {H^\Theta } = + 50.6{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}} \\ {{{\text{N}}_2}{{\text{H}}_4}({\text{l)}} \to {{\text{N}}_2}{{\text{H}}_4}({\text{g)}}}&{\Delta {H^\Theta } = + 44.8{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}} \end{array}$$

What is ${\Delta {H^\Theta }}$, in kJ, for the following reaction?

$${{\text{N}}_2}({\text{g)}} + 2{{\text{H}}_2}({\text{g)}} \to {{\text{N}}_2}{{\text{H}}_4}({\text{g)}}$$

A.     $- 95.4$

B.     $- 5.80$

C.     $+ 5.80$

D.     $+ 95.4$

D

There were two G2 comments on this question, with both saying that the question was too difficult for SL candidates, especially without the use of a calculator. This type of question has been asked on P1 several times before, and this in general was not an issue at all for candidates, with 74.00% of candidates getting the correct answer D. It is true that algebraic variables could have been used, though in this case the calculation involved is relatively simple: $+ 50.6 + ( + 44.8) =  + 95.4{\text{ kJ}}$ and is simply the addition of two numbers, since no equation inversion is involved nor is a multiplication factor necessary.