Question

What is the difference between ΔH and ΔU at 298 K for the following reaction?

\[\ce{C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l); \Delta H = - 1410.0 kJ}\]

Options

• - 14.8 kJ

• - 2.45 kJ

• - 4. 95 kJ

• - 7.30 kJ

Answer 1

- 4. 95 kJ

Explanation:

In the given reaction,

\[\ce{C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l)}\]

ΔH = - 1410.0 kJ

For ideal gas, ΔH = ΔU + RT(Δn_g)

where, R = 8.314 J/mol K

T = 298 K

Δn_g = number of gaseous products - number of gaseous reactants

= 2 - 4 = -2

∴ - 1410 = ΔU + [8.314 × 298 (- 2)]

ΔU = - 1405.045 kJ ≅ - 1405.05 kJ

Now, ΔH - ΔU = - 1410 - (- 1405.05) = - 4.95 kJ