Question

The heat of evaporation of water is 44.6 kJ mol−1 at 25℃. When 1000 g water vapour is condensed ∆H will be​

Answer 1

Explanation:

We know that for gaseous reactants and products , we have a relation between standard enthalpy of vapourization (ΔH

vap.

) and standard internal energy (ΔE) as-

ΔH

vap.

=ΔE+Δn

g

RT

whereas,

Δn

g

=n

2

−n

1

, i.e., difference between no. of moles of reactant and product.

For vapourization of water,

H

2

O

(l)

⟶H

2

O

(g)

∴Δn

g

=1−0=1

T=100℃=(100+273)=373K

∴ΔH

vap.

=ΔE+Δn

g

RT

⇒ΔE=ΔH

vap.

−Δn

g

RT=40.63−(1×8.314×10

−3

×373)=37.53KJ/mol

Hence the value ΔE for this process will be 37.53KJ/mol

Hence, the correct option is A.

Answer 2

Answer:

option A is correct .

Explanation:

Plz mark me as brain list