Q. Given, (i)$S + O_{2} \rightarrow SO_{2},\ \ \mathrm{\Delta}H = - 298.2$ kJ (ii)${SO}_{2} + \frac{1}{2}O_{2} \rightarrow SO_{3},\ \ \mathrm{\Delta}H = - 98.7$ kJ (iii)${SO}_{3} + H_{2}O \rightarrow H_{2}{SO}_{4},\ \ \mathrm{\Delta}H = - 130.2$ kJ (iv)$H_{2} + \frac{1}{2}O_{2} \rightarrow H_{2}O,\ \ \mathrm{\Delta}H = - 287.3$ kJ Then the enthalpy of formation of $H_{2}SO_{4}$ at 298 K will be
A) $- 814.4$ kJ
B) $+ 320.5$ kJ
C) $- 650.3$ kJ
D) $- 933.7$ kJ
Explanation
(a)
For the equation,
$H_{2} + S + 2O_{2} \rightarrow H_{2}SO_{4}$
Eqs. (i) +(ii)+(iii)+(iv)
$\mathrm{\Delta}H = - 287.3 + (298.2) + ( - 98.7) + ( - 130.2)$
$= 814.4\ kJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.