For N2O4(g) -> 2NO2(g) at 298 K, Delta H=+57.2 kJ per mole of N2O4. Taking R=8.314 J mol^-1K^-1, calculate Delta U in kJ mol^-1 (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3td132

For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ at $298 K$ , $Δ H = + 57.2 kJ$ per mole of $N X_{2} O X_{4}$ . Taking $R = 8.314 J mo l^{- 1} K^{- 1}$ , calculate $Δ U$ in $kJ mo l^{- 1}$ (two decimals).

Numerical answer type, answer in kJ/mol, tolerance 0.02.

Δ U = Δ H - Δ n_{g} RT

with

Δ n_{g} = 2 - 1 = + 1

Δ n_{g} RT = 8.314 \times 298/1000 = 2.48 kJ

Δ U = 57.2 - 2.48 = 54.72 kJ mo l^{- 1}