Calculate Delta H for 4NH3 + 5O2 --> 4NO + 6H2O
using:
N2 +O2 --> 2NO DH= -180.5 kJ
N2 + 3H2 --> 2NH3 DH= -91.8 kJ
2H2 + O2 --> 2H2O DH= -483.6 kJ
-1628.2
To calculate ΔH for the given reaction, we can use Hess's Law, which states that the enthalpy change for a reaction is the same regardless of the route taken.
First, we need to manipulate the given equations to match the desired reaction:
1) Reverse equation 1:
2NO --> N2 + O2 (ΔH = +180.5 kJ)
2) Multiply equation 2 by 2:
2N2 + 6H2 --> 4NH3 (ΔH = -183.6 kJ)
3) Multiply equation 3 by 3:
6H2 + 3O2 --> 6H2O (ΔH = -1450.8 kJ)
Now, we add the manipulated equations together to obtain the desired reaction:
2NO + 2N2 + 6H2 + 3O2 --> 2N2 + O2 + 4NH3 + 6H2O
Cancelling out the common species on both sides of the equation:
2NO + 6H2 + 3O2 --> 4NH3 + 6H2O
Now, we add up the respective ΔH values of the manipulated equations to obtain the overall ΔH for the desired reaction:
(+180.5 kJ) + (-183.6 kJ) + (-1450.8 kJ) = -1453.9 kJ
Therefore, the ΔH for the given reaction 4NH3 + 5O2 --> 4NO + 6H2O is -1453.9 kJ.
To calculate the enthalpy change (ΔH) for the reaction 4NH3 + 5O2 → 4NO + 6H2O using the given equations, you need to use a combination of Hess's Law and stoichiometry.
1. Rearrange the given equations to match the reactants and products in the desired reaction:
a) N2 + O2 → 2NO
b) N2 + 3H2 → 2NH3
c) 2H2 + O2 → 2H2O
2. Multiply the relevant equations to obtain the desired reaction equation:
a) 2 * equation a: 2(N2 + O2 → 2NO) gives 2N2 + 2O2 → 4NO
b) equation b: N2 + 3H2 → 2NH3
c) 3 * equation c: 3(2H2 + O2 → 2H2O) gives 6H2 + 3O2 → 6H2O
3. Combine the equations obtained in step 2 to construct the desired reaction equation:
4NH3 + 5O2 → 4NO + 6H2O
4. Sum the enthalpy changes (ΔH) of the equations to calculate the overall ΔH for the desired reaction.
a) 2 * equation a: ΔH = 2 * -180.5 kJ = -361 kJ
b) equation b: ΔH = -91.8 kJ
c) 3 * equation c: ΔH = 3 * -483.6 kJ = -1451 kJ
5. Add up the individual enthalpy changes calculated in step 4 to find the overall ΔH:
ΔH = (-361 kJ) + (-91.8 kJ) + (-1451 kJ) = -1903.8 kJ
Therefore, the ΔH for the reaction 4NH3 + 5O2 → 4NO + 6H2O is -1903.8 kJ.
Hess' Law.
Multiply equation 2 by 2 and reverse it.
Multiply equation 3 by 3 and leave as is.
Multiply equation 1 by 2 and leave as is.
Add the resulting equations.
Add the resulting delta Hs also, (after multipling AND changing the sign of those reversed.