Question

Standard enthalpy of vaporisation Δvap H⁰ for water at 100⁰C is 40.66 kJ mol⁻¹. The internal energy of vaporisation of water at 100⁰C (in kJ mol⁻¹) is

Answer 1

Explanation:

Q = msΔt , Q = 1.5 x 4.2 x (60-30) x 1000, Q = 189 x 10³ J , Calorific value = 189 x 10³ / 6.3 = 30 kJ/g

Answer 2

QUESTION

• Standard enthalpy of vaporisation Δvap H⁰ for water at 100⁰C is 40.66 kJ mol⁻¹. The internal energy of vaporisation of water at 100⁰C (in kJ mol⁻¹) is

EXPLANATION

GiVEN

• ΔvapH⁰ = 40.66 kJ mol⁻¹
• T = 100 + 273 = 373K

TO FiND

• ΔE = ¿?

SOLUTiON

I know that

ΔH = ΔE + Δn(g)RT

ΔE = ΔH – Δn(g) RT

where

• Δn(g) = number of gaseous moles of products – number of gaseous moles of reactants

H₂O(l) ⇌ H₂O(g)

So, Δn(g) = 1 – 0 = 1

ΔE = ΔH - RT

ΔE = ( 40.66 X 10³) – (8.314 X 373)

ΔE = 37.56 kJ/mol