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
To calculate the enthalpy change for the given reaction, we can use the Hess's Law.
Let's break down the reaction into multiple steps using the given thermochemical equations:
1) Multiply the second equation by 2 to balance the nitrogen (N2) on the reactant side:
2 N2 (g) + 6 H2 (g) → 4 NH3 (g) ΔrH° = -183.6 kJ
2) Multiply the third equation by 2 to balance the water (H2O) on the product side:
4 H2 (g) + 2 O2 (g) → 4 H2O (g) ΔrH° = -968 kJ
3) Multiply the first equation by 4 to balance the nitrogen monoxide (NO) on the product side:
2 N2 (g) + 4 O2 (g) → 8 NO (g) ΔrH° = +724 kJ
Now, we can add these three equations to obtain the overall reaction:
2 N2 (g) + 4 O2 (g) + 4 H2 (g) → 8 NO (g) + 4 H2O (g) + 4 NH3 (g)
The enthalpy change for the overall reaction is obtained by adding the enthalpy changes of the individual reactions:
ΔrH = ΔrH1 + ΔrH2 + ΔrH3
= (-183.6 kJ) + (-968 kJ) + (+724 kJ)
= -1427.6 kJ
Therefore, the enthalpy change for the given reaction is -1427.6 kJ.
To calculate the enthalpy change (ΔrH) for the given reaction, you can use the concept of Hess's Law. Hess's Law states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states.
To calculate ΔrH for the given reaction, you can manipulate the given thermochemical equations in a way that the reactants and products match those in the given reaction. Here's a step-by-step approach:
Step 1: Reverse the second equation and double the first equation to match the number of moles of NO and NH3 in the given reaction.
- (1) 2 N2 (g) + 2 O2 (g) → 4 NO (g) ΔrH° = +362 kJ
- (2) 2 N2 (g) + 6 H2 (g) → 4 NH3 (g) ΔrH° = -183.6 kJ
Step 2: Multiply the third equation by 3 to match the number of moles of H2O.
- (3) 6 H2 (g) + 3 O2 (g) → 6 H2O (g) ΔrH° = -1452 kJ
Step 3: Combine the manipulated equations to obtain the overall equation for the given reaction.
- Add equation (1) + equation (2) + equation (3):
2 N2 (g) + 2 O2 (g) + 2 N2 (g) + 6 H2 (g) + 6 H2 (g) + 3 O2 (g) →
4 NO (g) + 4 NH3 (g) + 6 H2O (g)
Simplify to get:
4 N2 (g) + 9 O2 (g) + 6 H2 (g) → 4 NO (g) + 4 NH3 (g) + 6 H2O (g)
Step 4: Sum up the ΔrH° values for each equation to obtain the ΔrH° for the overall reaction.
ΔrH° (overall) = ΔrH°(1) + ΔrH°(2) + ΔrH°(3)
= +362 kJ + (-183.6 kJ) + (-1452 kJ)
= -1273.6 kJ
Therefore, the enthalpy change (ΔrH) for the given reaction is -1273.6 kJ.