Nitrogen reacts with hydrogen to form ammonia according to the equation below.
mc023-1.jpg
Standard Enthalpies of Formation
Substance
DHf (kJ/mol)
C2 H2 (g)
–26.7
NH3 (g)
–46.19
HBr (g)
236.23
HCl (g)
–92.30
HF (g)
–268.6
HI (g)
25.9
NaCl (s)
–411.0
Based on the equation and the information in the table, what is the enthalpy of the reaction?
mc023-2.jpgHrxn = 46.19 kJ
mc023-3.jpgHrxn = –46.19 kJ
mc023-4.jpgHrxn = 92.38 kJ
mc023-5.jpgHrxn = –92.38 kJ
To calculate the enthalpy of the reaction, we need to determine the difference in enthalpies of formation between the products and the reactants.
ΔHrxn = ΣnΔHf(products) - ΣmΔHf(reactants)
In this case, the products are NH3, and the reactants are N2 and H2.
NH3: ΔHf = -46.19 kJ/mol
N2: ΔHf = 0 (since it's in its standard state)
H2: ΔHf = 0 (since it's in its standard state)
ΔHrxn = [1(-46.19)] - [1(0) + 3(0)]
ΔHrxn = -46.19 kJ
Therefore, the enthalpy of the reaction is -46.19 kJ. This means that the reaction is exothermic. So the correct answer is:
mc023-3.jpgΔHrxn = –46.19 kJ