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Question
If 38.55 kJ of heat is absorbed, when 6.0 of O2 react CIF according to reaction.
\[\ce{2CIF_{(g)} + O2_{(g)} -> Cl2_{(g)} + OF2_{(g)}}\]
What is the standard enthalpy of reaction?
Options
102.8 kJ
49.80 kJ
72.28 kJ
205.6 kJ
MCQ
Solution
205.6 kJ
Explanation:
Given,
Enthalpy change for a given mass = 38.55 kJ
Mass of O2 = 6.0 g
Number of moles of O2 = `("Mass of O"_2)/("Molar mass of O"_2)`
= `(6 "g")/(32 "g mol"^-1)`
= 0.1875 mol
∴ Enthalpy change for 1 mole O2 = `(38.55 "k"J)/(0.1875)`
= 205.6 kJ
2 moles of CIF react with 1 mole of O2 in this reaction.
As a result, the usual reaction enthalpy is + 205.6 kJ.
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Nature of Heat and Work
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