If enthalpies of formation of C_2 H_4( g), CO_2( g) and H_2 O(l) at 25^ ^ C and 1 atm pressure are 52,394 and -286 kJ / mol respectively, the change in enthalpy for combustion of C_2 H_4 is equal to

Correct answer: B. -1412 kJ / mol Enthalpy of formation of C_2 H_4, CO_2 and H_2 O are 52,-394 and -286 kJ / mol respectively. (Given) The reaction is C_2 H_4+3 O_2 arrow 2 CO_2+2 H_2 O change in enthalpy, (Δ H)=Δ H_ products -Δ H_ reactants =2 ×(-394)+2 ×(-286)-(52+0) =-1412 kJ / mol .

BITSAT 2023Paper 1

Exam: BITSAT
Year: 2023
Paper: Paper 1
Subject: Chemistry
Chapter: Thermodynamics (C)

BITSAT 2023 · Paper 1 · Chemistry · Thermodynamics (C)

If enthalpies of formation of C_2 H_4( g), CO_2( g) and H_2 O(l) at...

If enthalpies of formation of

\mathrm{C}_2 \mathrm{H}_4(\mathrm{ g}), \mathrm{CO}_2(\mathrm{ g})

and

\mathrm{H}_2 \mathrm{O}(\mathrm{l})

25^{{}^{\circ}} \mathrm{C}

and

1 \mathrm{ atm}

pressure are 52,394 and

-286 \mathrm{ kJ} / \mathrm{mol}

respectively, the change in enthalpy for combustion of

\mathrm{C}_2 \mathrm{H}_4

is equal to

Options

A $-141.2 \mathrm{ kJ} / \mathrm{mol}$
B $-1412 \mathrm{ kJ} / \mathrm{mol}$
C $+14.2 \mathrm{ kJ} / \mathrm{mol}$
D $+1412 \mathrm{ kJ} / \mathrm{mol}$

Correct answer

B. $-1412 \mathrm{ kJ} / \mathrm{mol}$

Step-by-step solution

Enthalpy of formation of

\mathrm{C}_2 \mathrm{H}_4, \mathrm{CO}_2

and

\mathrm{H}_2 \mathrm{O}

are

52,-394

and

-286 \mathrm{ kJ} / \mathrm{mol}

respectively.
(Given)
The reaction is

\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}

change in enthalpy,

(\Delta H)=\Delta H_{\text{products }}-\Delta H_{\text{reactants }}

=2 \times(-394)+2 \times(-286)-(52+0)

=-1412 \mathrm{ kJ} / \mathrm{mol}

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