The latent heat of vapourisation of water is 44.0 kJ mol-1. Calculate the heat energy, in kJ, required to evaporate 200mg of water at 100 degrees Celcius.
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Solution:- (A) 37.53KJ/mol
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.
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