Question
The enthalpy of formation of ammonia gas is -46kJ \(mol^{-1}\)
\(N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)\)
What is the energy released, in KJ, in the reaction?
A. 23
B. 46
C. 69
D. 92
▶️Answer/Explanation
Markscheme: D
The enthalpy of formation (\(\Delta H_f\)) of a reaction is the heat released or absorbed when one mole of a compound is formed from its elements in their standard states.
In this case, the enthalpy of formation of ammonia (\(NH_3\)) is given as -46 kJ/mol. The balanced chemical equation for the formation of ammonia from nitrogen (\(N_2\)) and hydrogen (\(H_2\)) is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
To find the energy released in the reaction, you need to multiply the enthalpy of formation by the stoichiometric coefficient for the product. In this case, the coefficient for \(NH_3\) is 2. Therefore:
\[ \text{Energy released} = 2 \times \text{Enthalpy of formation of } NH_3 \]
\[ \text{Energy released} = 2 \times (-46 \, \text{kJ/mol}) \]
\[ \text{Energy released} = -92 \, \text{kJ} \]
The negative sign indicates that the reaction is exothermic, releasing energy.
Question
What is the enthalpy change for the following reaction?
A. -74 – 242 + 111
B. +74 + 242 + 111
C. -74 + 242 – 111
D. +74 + 484 – 222
Answer/Explanation
Markscheme: B