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.