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Sagot :
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
-----------------------------------------------------------------------------------------
2NH3(g) + 2CH4(g) ⇒ 2HCN + 6H2 ∆ H: 511.9 ⇒ :2
NH3(g) + CH4(g) ⇒ HCN + 3H2 ∆H:255.95
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