Question

the enthalpy of the formation of CO2 & H2O are -395KJ &-285KJ respectively and the enthalpy of combustion of acetic acidis 869KJ. The enthalpy of acetic acid is?​

Answer 1

Answer:

491.8kJ/mol

Explanation:

Combustion reaction for above question is

C

2

H

4

O

2

+

2

O

2

→

2

C

O

2

+

2

H

2

O

Δ

H

f

(

c

o

m

b

u

s

t

i

o

n

)

=

2

[

Δ

H

C

O

2

+

Δ

H

H

2

O

]

−

Δ

H

C

2

H

4

O

2

Δ

H

C

2

H

4

O

2

=

2

[

Δ

H

C

O

2

+

Δ

H

H

2

O

]

−

Δ

H

c

o

m

b

u

s

t

i

o

n

Δ

H

C

2

H

4

O

2

=

867

+

(

2

×

−

679.4

)

=

−

491.8

k

J

m

o

l

Answer 2

ΔH(combustion) = -860kJ/mol

ΔH(CO2) = -393.5kJ/mol

ΔH(H20) = -285.9kJ/mol

# Solution-

Combustion reaction occurs as follows,

C2H4O2 + 2O2 --> 2CO2 + 2H2O

Enthalpy of combustion is given by,

ΔH(combustion) = 2ΔH(CO2) + 2ΔH(H2O) -ΔH(C2H4O2)

ΔH(C2H4O2) = 2ΔH(CO2) + 2ΔH(H2O) - ΔH(combustion)

ΔH(C2H4O2) = 2×(-393.5) + 2×(-285.9) - (-860)

ΔH(C2H4O2) = -498.8 kJ/mol

# Enthalpy of formation of acetic acid is -498.8 kJ/mol.

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