If,${\rm{S}} + {{\rm{O}}_2} \to {\rm{S}}{{\rm{O}}_2};∆H = - 298.2{\rm{kJ}}$...(i)${\rm{S}}{{\rm{O}}_2} + \frac{1}{2}{{\rm{O}}_2} \to {\rm{S}}{{\rm{O}}_3};∆H = - 98.7{\rm{kJ}}$...(ii)${\rm{S}}{{\rm{O}}_3} + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4};∆H = - 130.2{\rm{kJ}}$...(iii)${{\rm{H}}_2} + \frac{1}{2}{{\rm{O}}_2} \to {{\rm{H}}_2}{\rm{O}};∆H = - 227.3{\rm{kJ}}$...(iv)The enthalpy of formation of ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$ at 298 K will be:

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

1 of 1

Application +4 | -1

If,

\mathrm{S} + {\mathrm{O}_2} \to \mathrm{S}{\mathrm{O}_2};∆H = - 298.2\mathrm{kJ}

...(i)

\mathrm{S}{\mathrm{O}_2} + \frac{1}{2}{\mathrm{O}_2} \to \mathrm{S}{\mathrm{O}_3};∆H = - 98.7\mathrm{kJ}

...(ii)

\mathrm{S}{\mathrm{O}_3} + {\mathrm{H}_2}\mathrm{O} \to {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4};∆H = - 130.2\mathrm{kJ}

...(iii)

{\mathrm{H}_2} + \frac{1}{2}{\mathrm{O}_2} \to {\mathrm{H}_2}\mathrm{O};∆H = - 227.3\mathrm{kJ}

...(iv)The enthalpy of formation of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

at 298 K will be:

View answer and explanation

Correct option: A

- 754.4\mathrm{kJ}

Option A (-754.4 kJ) is correct because it represents the sum of the enthalpy changes of the reactions that form H2SO4 from its elements.