Question

ΔH for the formation of ethane gas is −84.4 kJ at 300 K. Calculate ΔU for the reaction.

Answer 1

Given:

ΔH = −84.4 kJ 

Temperature, T = 300 K

Gas constant, R = 8.314 J/mol·K = 0.008314 kJ/mol·K

Δn_g​ is the change in the number of gaseous moles between products and reactants.

Formation reaction of ethane gas:

\[\ce{2C(s) + 3H2(g)->C2H6(g)}\]

Moles of gaseous reactants: 3 (from H₂)

Moles of gaseous products: 1 (from C₂H₆)

Δn_g ​= Moles of gaseous products − Moles of gaseous reactants

= 1 − 3

= −2

ΔU = ΔH − Δn_g​RT

ΔU = (−84.4) − (−2 × 0.008314 × 300)

ΔU = −84.4 + (2 × 2.4942)

ΔU = −84.4 + 4.9884

ΔU = −79.41 kJ