Question

Can someone please walk me through this?-- 25 pts!
Calculate the change of enthalpy for the reaction CH4 (g) + NH3 (g) --> HCN (g) +3H2 (g) from the following reactions:

Reaction 1: N2 (g) + 3H2 (g) --> 2NH3 (g); Change in enthalpy: -91.8 kJ/mol

Reaction 2: C (s, graphite) + 2H2 (g) --> CH4 (g); Change in enthalply: -74.9 kJ/mol

Reaction 3: H2 (g) + 2C (s, graphite) + N2 (g) --> 2HCN (g); Change in enthalpy: +270.3 kJ/mol

I have to include the following:

The numerical answer with correct units.
State which reactions, if any, you had to "Flip".
State which reactions you had to multiply, if any, to get the correct amount of the compound. Also, include how much you multiplied the reaction by.

Answer 1

NH3(g) + CH4(g) ⇒ HCN + 3H2 ∆ H:255.95  kJ/mol

Further explanation

Given

Reaction and the enthalpy

Required

The change of enthalpy

Solution

Reaction 1

N2 (g) + 3H2 (g) --> 2NH3 ∆ H: -91.8 kJ/mol⇒reverse

2NH3 ⇒ N2 (g) + 3H2 (g) ∆ H: +91.8 kJ/mol

Reaction 2

C (s, graphite) + 2H2 (g)⇒ CH4 (g) ∆H: -74.9 kJ/mol ⇒reverse

CH4 (g) ⇒ C (s, graphite) + 2H2 (g) ∆H: +74.9 kJ/mol ⇒ x2

2CH4 (g) ⇒ 2C (s, graphite) + 4H2 (g) ∆H: +149.8 kJ/mo

Reaction 3

H2 (g) + 2C (s, graphite) + N2 (g) ⇒ 2HCN (g);∆ H: +270.3 kJ/mol

Add up all the reactions and remove the same compound from different sides :

2NH3 ⇒ N2 (g) + 3H2 (g) ∆ H: +91.8 kJ/mol

2CH4 (g) ⇒ 2C (s, graphite) + 4H2 (g) ∆H: +149.8 kJ/mol

H2 (g) + 2C (s, graphite) + N2 (g) ⇒ 2HCN (g) ∆ H: +270.3 kJ/mol

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2NH3(g) + 2CH4(g) ⇒ 2HCN + 6H2 ∆ H: 511.9 ⇒ :2

NH3(g) + CH4(g) ⇒ HCN + 3H2 ∆H:255.95