The ∆H θ for the reaction, 4S(s) + 6O 2 (g) → 4 SO 3 (g) is –1583.2 kJ. Standard enthalpy…
Chemistry · JEE Main · NTA Exams — Chemical Thermodynamics
The ∆Hθ for the reaction,
4S(s) + 6O2 (g)\(\text { → }\)4 SO3(g)
is –1583.2 kJ. Standard enthalpy of formation of sulphur trioxide is :
4S(s) + 6O2 (g)\(\text { → }\)4 SO3(g)
is –1583.2 kJ. Standard enthalpy of formation of sulphur trioxide is :
- –3166.4 kJ
- 3166.4 kJ
- –395.8 kJ
- 395.8 kJ
Answer
(C) –395.8 kJ
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