Calculate the enthalpy of formation of methanol from the following
data: (4)
CH3OH + 3/2 O2 → CO2 + 2H2O △rH
o=-726 kJ/mol
C + O2 → CO2 △cH
o=-393 kJ/mol
H2 + ½ O2 → H2O △fH
o=-286 kJ/mol
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Answer:
Explanation:. On reversing the first equation we get,
CO
2
(g)+2H
2
O(l)→CH
3
OH(l)+
2
3
O
2
(g); Δ
r
H
θ
=+726kJmol
−1
(on reversing the reaction, the sign of ΔH also reverses)
The second equation remains as such,
C
(graphite)
+O
2
(g)→CO
2
(g);Δ
c
H
θ
=−393kJmol
−1
On multiplying equation three by 2 we get,
2H
2
(g)+O
2
(g)→2H
2
O(l);2Δ
f
H
θ
=2×−286kJmol
−1
=−572kJmol
−1
On adding these three equations we get,
C
(graphite)
+2H
2
(g)+
2
1
O
2
(g)→CH
3
OH(l)
Enthalpy of formation of CH
3
OH=Δ
f
H=Δ
r
H
θ
+Δ
c
H
θ
+2Δ
f
H
θ
Δ
f
H=+726−393−572kJmol
−1
=−239kJmol
−1
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