Question

In: Chemistry

Hydrogen cyanide is a highly poisonous substance that can be prepared by the reaction of methane...

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

Solutions

Expert Solution

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


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