JEE Main 2025 · Chemistry · Chemical Thermodynamics · medium

Enthalpies of combustion at 298 K (products CO2(g) and H2O(l)) are: graphite -393.5, H2(g) -285.8, C2H6(g) -1560 kJ mol^-1. For the formation…

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Enthalpies of combustion at

298\ \mathrm{K}

(products

\mathrm{CO_2(g)}

and

\mathrm{H_2O(l)}

) are: graphite

-393.5

\mathrm{H_2(g)}

-285.8

\mathrm{C_2H_6(g)}

-1560\ \mathrm{kJ\,mol^{-1}}

. For the formation

2\mathrm{C(graphite)}+3\mathrm{H_2(g)}\to\mathrm{C_2H_6(g)}

, the change in internal energy

\Delta U

\mathrm{kJ\,mol^{-1}}

is closest to (

R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}

)

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Answer: (C) $-79.4$

\Delta_fH = 2(-393.5)+3(-285.8)-(-1560) = -787.0-857.4+1560 = -84.4\ \mathrm{kJ\,mol^{-1}}

. Gaseous moles change from

3

(only

\mathrm{H_2}

; graphite is solid) to

1

, so

\Delta n_g=-2

. Then

\Delta U=\Delta H-\Delta n_gRT = -84.4+2(8.314)(298)/1000 = -84.4+4.96 = -79.4\ \mathrm{kJ\,mol^{-1}}

. Using

\Delta n_g=-1

would give

-81.9

and ignoring it gives

-84.4

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