Calculate the ΔHo for the reaction, 6NO2(g) + 8NH3(g) → 7N2(g) + 12H2O(l).
N2(g) + 2O2(g) → 2NO2(g) ΔHo=+66.36 kJ
4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) ΔHo=-1530.36 kJ
-787.30 kJ
-3259.80 kJ
-1464.00 kJ
-2861.64 kJ
To calculate the ΔHo for the reaction 6NO2(g) + 8NH3(g) → 7N2(g) + 12H2O(l), we can use the given reactions and their enthalpy values.
We can start by scaling the given reactions so that the number of moles of the desired reaction matches the coefficients in the given reaction:
2 * (N2(g) + 2O2(g) → 2NO2(g))
4 * (NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l))
This gives us:
2N2(g) + 4O2(g) → 4NO2(g) ΔHo = 2 * (+66.36 kJ) = +132.72 kJ
8NH3(g) + 24O2(g) → 16N2(g) + 48H2O(l) ΔHo = 4 * (-1530.36 kJ) = -6121.44 kJ
Now, we can reverse the first reaction and multiply the second reaction by 7, so they cancel out:
-2 * (4NO2(g) → 2N2(g) + 4O2(g))
7 * (16N2(g) + 48H2O(l) → 8NH3(g) + 24O2(g))
This gives us:
-4NO2(g) + 2N2(g) + 4O2(g) ΔHo = -2 * (+66.36 kJ) = -132.72 kJ
112N2(g) + 336H2O(l) → 56NH3(g) + 168O2(g) ΔHo = 7 * (-6121.44 kJ) = -42850.08 kJ
Now, we can add the two reactions together to get the desired reaction:
-4NO2(g) + 2N2(g) + 4O2(g)
112N2(g) + 336H2O(l)
6NO2(g) + 8NH3(g) → 7N2(g) + 12H2O(l) ΔHo = (-132.72 kJ) + (-42850.08 kJ) = -42982.80 kJ
Therefore, the ΔHo for the reaction is -42982.80 kJ or -4.298280 x 10^4 kJ.