In: Chemistry
Hydrogen cyanide is a highly poisonous substance that can be prepared by the reaction of methane and ammonia:
CH4 (g) + NH3 (g) → HCN(g) + 3 H2 (g)
What is the heat of reaction at constant pressure? Provide an answer correct to 3 significant figures. Do not use scientific notation or include units to report your answer. You do need to include the sign as a part of your answer.
N2 (g) + 3 H2 (g) → 2 NH3
(g)
ΔHrxn = -91.8 kJ;
C(s) + 2 H2 (g) → CH4
(g)
ΔHrxn = -74.9 kJ;
2 C(s) + H2 (g) + N2 (g) → 2 HCN(g)
ΔHrxn = +270.3 kJ
Objective:
CH4 (g) + NH3 (g) → HCN(g) + 3 H2 (g)
From
N2 (g) + 3 H2 (g) → 2 NH3
(g)
ΔHrxn = -91.8 kJ;
C(s) + 2 H2 (g) → CH4
(g)
ΔHrxn = -74.9 kJ;
2 C(s) + H2 (g) + N2 (g) → 2 HCN(g)
ΔHrxn = +270.3 kJ
Invert (2)
N2 (g) + 3 H2 (g) → 2 NH3
(g)
ΔHrxn = -91.8 kJ;
CH4 (g) → C(s) + 2 H2 (g) ΔHrxn = +74.9
kJ;
2 C(s) + H2 (g) + N2 (g) → 2 HCN(g)
ΔHrxn = +270.3 kJ
invert (1)
2 NH3 (g)→ N2 (g) + 3 H2
(g)
ΔHrxn = +91.8 kJ;
CH4 (g) → C(s) + 2 H2 (g) ΔHrxn
= +74.9 kJ;
2 C(s) + H2 (g) + N2 (g) → 2 HCN(g)
ΔHrxn = +270.3 kJ
Divide (1) (this is tricky, the mass balance wont make sense, but the energy balance does...)
NH3 (g)→ 1/2N2 (g) + 3/2 H2
(g)
ΔHrxn = +91.8/2 kJ; = 45.9 kJ
CH4 (g) → C(s) + 2 H2 (g) ΔHrxn
= +74.9 kJ;
2 C(s) + H2 (g) + N2 (g) → 2 HCN(g)
ΔHrxn = +270.3 kJ
Add all
NH3 (g) + CH4 (g)+2 C(s) + H2 (g) + N2 (g) → 1/2N2 (g) + 3/2 H2 (g) + C(s) + 2 H2 (g) + 2 HCN(g)
Cancel common terms
NH3 (g) + CH4 (g)+ C(s) ) + 1/2N2 (g) → 2.5 H2 (g) + 2 HCN(g) H = 45.9 kJ+74.9 kJ;+270.3 kJ = 391.1 kJ
NOTE that actually...
Entahlpy of formation of C, N2, H2, is 0
which means that actually the reaction:
NH3 (g) + CH4 (g)+ C(s) ) + 1/2N2 (g) → 2.5 H2 (g) + 2 HCN(g) H = 391.1 kJ
Has the same enethalpy energy as:
NH3 (g) + CH4 (g) → 2 HCN(g) H = 391.1 kJ
Since enthalpy of 3H2 is 0, we can add this to the equation
NH3 (g) + CH4 (g) → 3H2 (g) + 2 HCN(g) H = 391.1 kJ
NOTE that the mass balance has NOTHING to do with the enthalpy of reaction