Question

The enthalpy of vaporization of water at hundred degree celsius is 40.6 3 kilo joule per mole the value of

Answer 1

Answer:

enthalpy of vaporization for water is 37.55KJ/mol

Explanation:

Standard enthalpy of vaporization( ΔvapH0)  and standard internal energy( ΔU0) is given as :

ΔvapH0 =  ΔU0 +  Δng RT

where,Δng is the difference between no. of moles of reactant and product.

Here, for vaporization of water ,

H2O (l) -> H2O (g)

So, change in no. of moles ,Δng = 1-0 =1

Thus,  ΔvapH0 =  ΔU0 + RT

ΔU0 = ΔvapH0 - RT  

= 44.66 - ( 8.314 x 10-3 KJ/Kmol  x 373K)

= 40.66 - 3.10 KJ/mol

= 37.55 KJ/mol

Answer 2

Answer:

I hope my answer will help you