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

JEE Main 2022 — Chemistry Physical Chemistry

2022mcqmedium

At $25^{\circ}C$ and $1$ atm pressure, the enthalpies of combustion are as given below:

Substance	${H}_{2}$	$C$ (graphite)	${C}_{2}{H}_{6}(g)$
$\frac{{\Delta }_{c}{H}^{\Theta }}{{\mathrm{kJmol}}^{-1}}$	$-286.0$	$-394.0$	$-1560.0$

The enthalpy of formation of ethane is

Official previous-year question

Held on 24 Jun 2022 · Verified 6 Jul 2026.

Options

A
$+54.0\mathrm{kJ}{\mathrm{mol}}^{-1}$
B
$-68.0\mathrm{kJ}{\mathrm{mol}}^{-1}$
C
$-86.0\mathrm{kJ}{\mathrm{mol}}^{-1}$
D
$+97.0\mathrm{kJ}{\mathrm{mol}}^{-1}$

Did you get this right?

Solution

Given

(i) ${C}_{2}{H}_{6}(g)+\frac{7}{2}{O}_{2}(g)\longrightarrow 2{\mathrm{CO}}_{2}(g)+3{H}_{2}O(l){\Delta H}_{\mathrm{comb}}^{^{\circ}}=-1560.0\mathrm{kJ}/\mathrm{mole}$

(ii) $C(s)+{O}_{2}(g)\longrightarrow {\mathrm{CO}}_{2}(g){\Delta H}_{\mathrm{comb}}^{^{\circ}}=-394.0\mathrm{kJ}/\mathrm{mole}$

(ii) ${H}_{2}(g)+\frac{1}{2}{O}_{2}(g)\longrightarrow {H}_{2}O(g){\Delta H}_{\mathrm{comb}}^{^{\circ}}=-286.0\mathrm{kJ}/\mathrm{mole}$

Target $2C(s)+3{H}_{2}(g)\longrightarrow {C}_{2}{H}_{6}(g){\Delta H}_{\mathrm{rx}}^{^{\circ}}={\Delta H}_{f}^{^{\circ}}({C}_{2}{H}_{6},g)$

${\Delta H}_{r}^{^{\circ}}={\Delta H}_{c}^{^{\circ}}(\mathrm{reactant})-{\Delta H}_{c}^{^{\circ}}(\mathrm{Product})$

$=2\times (-394)+3(-286)-(-1560)$

$=-788-858+1560$

$=-86.0\mathrm{KJ}/\mathrm{mole}$

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JEE Main 2022 — Chemistry Physical Chemistry

2022mcqmedium

At $25^{\circ}C$ and $1$ atm pressure, the enthalpies of combustion are as given below:

Substance	${H}_{2}$	$C$ (graphite)	${C}_{2}{H}_{6}(g)$
$\frac{{\Delta }_{c}{H}^{\Theta }}{{\mathrm{kJmol}}^{-1}}$	$-286.0$	$-394.0$	$-1560.0$

The enthalpy of formation of ethane is

Official previous-year question

Held on 24 Jun 2022 · Verified 6 Jul 2026.

Options

A
$+54.0\mathrm{kJ}{\mathrm{mol}}^{-1}$
B
$-68.0\mathrm{kJ}{\mathrm{mol}}^{-1}$
C
$-86.0\mathrm{kJ}{\mathrm{mol}}^{-1}$
D
$+97.0\mathrm{kJ}{\mathrm{mol}}^{-1}$

Did you get this right?

Solution

Given

(i) ${C}_{2}{H}_{6}(g)+\frac{7}{2}{O}_{2}(g)\longrightarrow 2{\mathrm{CO}}_{2}(g)+3{H}_{2}O(l){\Delta H}_{\mathrm{comb}}^{^{\circ}}=-1560.0\mathrm{kJ}/\mathrm{mole}$

(ii) $C(s)+{O}_{2}(g)\longrightarrow {\mathrm{CO}}_{2}(g){\Delta H}_{\mathrm{comb}}^{^{\circ}}=-394.0\mathrm{kJ}/\mathrm{mole}$

(ii) ${H}_{2}(g)+\frac{1}{2}{O}_{2}(g)\longrightarrow {H}_{2}O(g){\Delta H}_{\mathrm{comb}}^{^{\circ}}=-286.0\mathrm{kJ}/\mathrm{mole}$

Target $2C(s)+3{H}_{2}(g)\longrightarrow {C}_{2}{H}_{6}(g){\Delta H}_{\mathrm{rx}}^{^{\circ}}={\Delta H}_{f}^{^{\circ}}({C}_{2}{H}_{6},g)$

${\Delta H}_{r}^{^{\circ}}={\Delta H}_{c}^{^{\circ}}(\mathrm{reactant})-{\Delta H}_{c}^{^{\circ}}(\mathrm{Product})$

$=2\times (-394)+3(-286)-(-1560)$

$=-788-858+1560$

$=-86.0\mathrm{KJ}/\mathrm{mole}$

Your note

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JEE Main 2022 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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