At $25^{\circ}\mathrm{C}$ and 1 atm pressure, the enthalpies of combustion are as given below: | Substance | $\mathrm{H}_{2}$ | C(graphite) | $\mathrm{C}_{2}\mathrm{H}_{6}$(g) | |-----------|---------------------|-------------|---------------------------------------| | $\frac{\Delta_{\mathrm{C}}\mathrm{H}^{\Theta}}{\mathrm{kJmol}^{-1}}$ | -286.0 | -394.0 | -1560.0 | The enthalpy of formation of ethane is

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Chemical Thermodynamics

2 5^{\circ} C

and 1 atm pressure, the enthalpies of combustion are as given below:

Substance	$H_{2}$	C(graphite)	$C_{2} H_{6}$ (g)
$\frac{Δ _{C} H ^{Θ}}{kJmol ^{- 1}}$	-286.0	-394.0	-1560.0

The enthalpy of formation of ethane is

Answer: (C)

For the combustion of ethane:

C_{2} H_{6} (g) + \frac{7}{2} O_{2} (g) \to 2 CO_{2} (g) + 3 H_{2} O (ℓ)

Using Hess's law:

Δ_{C} H (C_{2} H_{6}) = 2 Δ_{f} H (CO_{2}) + 3 Δ_{f} H (H_{2} O) - Δ_{f} H (C_{2} H_{6})

- 1560 = 2 (- 394) + 3 (- 286) - Δ_{f} H (C_{2} H_{6})

- 1560 = - 788 - 858 - Δ_{f} H (C_{2} H_{6})

Δ_{f} H (C_{2} H_{6}) = - 86

kJ/mol

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option