Question

Write equations to express enthalpies of formation of
(a) Sulphur trioxide (c) water
(b) Ethyl alcohol
Plzz help.. ​

Answer 1

Answer:

ethyl alcohol, this is the answer

Answer 2

Answer:

C

2

H

5

OH

(l)

+3O

2

(g)

⟶2CO

2

(g)

+3H

2

O

(l)

ΔH=−1368kJ

2CO

2

(g)

+3H

2

O

(l)

⟶C

2

H

5

OH

(l)

+3O

2

(g)

ΔH=1368kJ.....(1)

C

(s)

+O

2

(g)

⟶CO

2

(g)

ΔH=−393.5kJ

2×[C

(s)

+O

2

(g)

⟶CO

2

(g)

ΔH=−393.5kJ]

2C

(s)

+2O

2

(g)

⟶2CO

2

(g)

ΔH=−787kJ.....(2)

H

2

(g)

+

2

1

O

2

(g)

⟶H

2

O

(l)

ΔH=−286.0kJ

3×[H

2

(g)

+

2

1

O

2

(g)

⟶H

2

O

(l)

ΔH=−286.0kJ]

3H

2

(g)

+

2

3

O

2

(g)

⟶3H

2

O

(l)

ΔH=−858.0kJ.....(3)

Adding eq

n

(1),(2)&(3), we have

2C

(s)

+3H

2

(g)

+

2

1

O

2

(g)

⟶C

2

H

5

OH

(l)

ΔH=−277

Hence the enthalpy of formation of ethyl alcohol is −277kJ.