The standard internal energy of combustion of C2H6(g) at 298 K is -1555.0 kJ mol^-1 (products CO2(g) and H2O(l)). Given Delta_f H deg(CO2,g) = -393.5 and Delta_f H deg(H2O,l) = -285.8 kJ mol^-1, and R = 8.314 J mol^-1K^-1 with ideal gases, calculate Delta_f H deg(C2H6,g) in kJ mol^-1 (one decimal place).

JEE Advanced levelNumerical answerDifficulty 3 of 3td012

The standard internal energy of combustion of $C X_{2} H X_{6} (g)$ at $298 K$ is $- 1555.0 kJ mo l^{- 1}$ (products $CO X_{2} (g)$ and $H X_{2} O (l)$ ). Given $Δ_{f} H^{\circ} (CO X_{2}, g) = - 393.5$ and $Δ_{f} H^{\circ} (H X_{2} O, l) = - 285.8 kJ mo l^{- 1}$ , and $R = 8.314 J mo l^{- 1} K^{- 1}$ with ideal gases, calculate $Δ_{f} H^{\circ} (C X_{2} H X_{6}, g)$ in $kJ mo l^{- 1}$ (one decimal place).

Numerical answer type, answer in kJ/mol, tolerance 0.1.

For

C X_{2} H X_{6} + 3.5 O X_{2} 2 CO X_{2} + 3 H X_{2} O (l)

Δ n_{g} = 2 - 4.5 = - 2.5

Δ_{c} H = - 1555.0 + (- 2.5) (8.314) (298) /1000 = - 1561.19

. Then

Δ_{c} H = 2 (- 393.5) + 3 (- 285.8) - Δ_{f} H (C X_{2} H X_{6})

, so

Δ_{f} H (C X_{2} H X_{6}) = - 1644.4 + 1561.19 = - 83.21 kJ mo l^{- 1}