Question

At a pressure of 1 atm, what is ^H in kj for the process of condensing a 42.5g sample of gaseous benzene at its normal boiling point of 80.1•c

Given C6H6
Boiling point 80.1
^Hvap: 30.7
Specific heat: 1.74

Answer 1

Answer:

-17.2 kJ

Step-by-step explanation:

For the process of vaporization,

C₆H₆(ℓ) ⇌ C₆H₆(g), ΔH(vap) = 30.7 kJ·mol⁻¹

For the process of condensation,

C₆H₆(g) ⇌ C₆H₆(ℓ), ΔH(cond) = -30.7 kJ·mol⁻¹

(a) Moles of benzene

n = 42.5 g × (1 mol/78.11 g)

  = 0.5441 mol

(b) Heat evolved

q = nΔH

  = 0.5441 mol × (-31.7 kJ/1 mol)

  = -17.2 kJ

For the condensation, ΔH = -17.2 kJ.