Calculate the enthalpy change, ΔrH, for the following reaction,
4 NH3 (g) + 5 O2(g) → 4 NO (g) + 6 H2O (g)
given the thermochemical equations below.
N2 (g) + O2 (g) → 2 NO (g) ΔrH° = +181 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔrH° = 91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) ΔrH° = 484 kJ
A. 127 kJ
B. 395 kJ
C. 906 kJ
D. 211 kJ
E. +1270 kJ
To calculate the enthalpy change (ΔrH) for the given reaction, we need to use Hess's Law, which states that the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes of the individual steps that make up the reaction.
In this case, we can break down the overall reaction into three steps using the given thermochemical equations:
Step 1: 4 NH3(g) → N2(g) + 3 H2(g) (reverse the second equation)
Step 2: N2(g) + O2(g) → 2 NO(g) (first equation)
Step 3: 2 H2(g) + O2(g) → 2 H2O(g) (third equation)
The enthalpy change for each step is given as follows:
ΔrH1 = -(-91.8 kJ) (Since we are reversing the second equation, we also reverse the sign of the enthalpy change)
ΔrH2 = +181 kJ
ΔrH3 = -484 kJ
To calculate ΔrH for the overall reaction, we need to add the enthalpy changes of the individual steps while taking into account the stoichiometric coefficients. The coefficients of the steps will be multiplied by the enthalpy changes of those steps:
ΔrH = (4 × ΔrH1) + (5 × ΔrH2) + (6 × ΔrH3)
Substituting the values:
ΔrH = (4 × -(-91.8 kJ)) + (5 × 181 kJ) + (6 × -484 kJ)
= (4 × 91.8 kJ) + (5 × 181 kJ) + (6 × -484 kJ)
= 367.2 kJ + 905 kJ - 2904 kJ
= -1631.8 kJ
The calculated enthalpy change (ΔrH) for the reaction is -1631.8 kJ. However, none of the given options match this value. Hence, none of the options provided (A, B, C, D, E) are correct.