If S+02→SO2; ∆H=-3982 kJ; SO2+12O2→SO3; ∆H=-98.7kJ SO3+H2O→H2SO4; ∆H=-130.2kJ; H2+12O2→H2O; ∆H=-227.3kJThe enthalpy of formation of sulphuric acid at 298K will be

S+O2→SO2 ΔH=−398.2kJ…..(1)SO2+12O2→SO3 ΔH=−98.7kJ…..2SO3+H2O→H2SO4 ΔH=−130.2kJ.....3H2+12O2→H2O ΔH=−227.3kJ….4H2+S+2O2→H2SO4 ΔH=−854.4kJ

If S+02→SO2; ∆H=-3982 kJ; SO2+12O2→SO3; ∆H=-98.7kJ SO3+H2O→H2SO4; ∆H=-130.2kJ; H2+12O2→H2O; ∆H=-227.3kJThe enthalpy of formation of sulphuric acid at 298K will be

S+O2→SO2 ΔH=−398.2kJ…..(1)SO2+12O2→SO3 ΔH=−98.7kJ…..2SO3+H2O→H2SO4 ΔH=−130.2kJ.....3H2+12O2→H2O ΔH=−227.3kJ….4H2+S+2O2→H2SO4 ΔH=−854.4kJ

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$If S + 0_{2} \to {SO}_{2}; ∆ H = - 3982 kJ; {SO}_{2} + \frac{1}{2} O_{2} \to {SO}_{3}; ∆ H = - 98.7 kJ {SO}_{3} + H_{2} O \to H_{2} {SO}_{4}; ∆ H = - 130.2 kJ; H_{2} + \frac{1}{2} O_{2} \to H_{2} O; ∆ H = - 227.3 kJ$ The enthalpy of formation of sulphuric acid at 298K will be

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$- 854.4 kJ$

$- 754.4 kJ$

$- 650.3 kJ$

$- 433.7 kJ$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{aligned} S + O_{2} \to {SO}_{2} ΔH = - 398.2 kJ \dots . . (1) \\ {SO}_{2} + \frac{1}{2} O_{2} \to {SO}_{3} ΔH = - 98.7 kJ \dots . . 2 \\ {SO}_{3} + H_{2} O \to H_{2} {SO}_{4} ΔH = - 130.2 kJ . . . . . 3 \\ H_{2} + \frac{1}{2} O_{2} \to H_{2} O ΔH = - 227.3 kJ \dots . 4 \end{aligned}$