Chemistry, asked by Anonymous, 3 months ago

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​

Answers

Answered by pratham7777775
2

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

c

H

θ

+2Δ

f

H

θ

Δ

f

H=+726−393−572kJmol

−1

=−239kJmol

−1

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