Given that standard heat enthalpy of ${\rm{C}}{{\rm{H}}_4},{{\rm{C}}_2}{{\rm{H}}_4}$and ${{\rm{C}}_3}{{\rm{H}}_8}$are −17.9,12.5,−24.8 kcal/mol. The ∆? for ${\rm{C}}{{\rm{H}}_4} + {{\rm{C}}_2}{{\rm{H}}_4} \to {{\rm{C}}_3}{{\rm{H}}_8}$ is :

$ - 19.4{\rm{kcal mo}}{{\rm{l}}^{ - 1}}$

$ - 55.2{\rm{kcal mo}}{{\rm{l}}^{ - 1}}$

$ - 30.2{\rm{kcal mo}}{{\rm{l}}^{ - 1}}$

$55.2{\rm{kcal mo}}{{\rm{l}}^{ - 1}}$

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NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1

Given that standard heat enthalpy of

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

and

{\mathrm{C}_3}{\mathrm{H}_8}

are −17.9,12.5,−24.8 kcal/mol. The ∆? for

\mathrm{C}{\mathrm{H}_4} + {\mathrm{C}_2}{\mathrm{H}_4} \to {\mathrm{C}_3}{\mathrm{H}_8}

is :

View answer and explanation

Correct option: D

- 19.4\mathrm{kcal mo}{\mathrm{l}^{ - 1}}

Option D (

- 19.4\mathrm{kcal mo}{\mathrm{l}^{ - 1}}

) is correct because it represents the enthalpy change calculated by subtracting the sum of the enthalpies of the reactants from the enthalpy of the product.

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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