Answer to Question #146113 in General Chemistry for nada

Question

Calculate ΔH for this reaction: CH4(g) + NH3(g) ---> HCN(g) + 3H2(g)
given:
N2(g) + 3H2(g) ---> 2NH3(g)
ΔH = −91.8 kJ
C(s) + 2H2(g) ---> CH4(g)
ΔH = −74.9 kJ
H2(g) + 2C(s) + N2(g) ---> 2HCN(g)
ΔH = +270.3 kJ

Answer 1

CH₄ : b reversible

CH₄(g) $\to$ C(graphite) + 2H₂(g)

∆H = + 74.9 KJ

NH₃ : a÷2 +reversible

NH₃(g) $\to$ $\frac{1}{2}$ N₂(g) + $\frac{3}{2}$ H₂(g)

∆H = + 45.9 KJ

HCN : c÷2 non reversible

$\frac{1}{2}$ H₂(g) +C(graphite) + $\frac{1}{2}$ N₂(g) $\to$ HCN(g)

∆H = +135.15KJ

CH₄(g) + NH₃(g) $\to$ HCN(g) +3H₂(g)

∆H = +74.9 KJ + 45.9KJ + 135.15KJ

∆H = +255.96KJ or

∆H = +256KJ