When 6.0 g of O2 reacts with CIF as per 2ClFA(g)+OA2A(g)⟶ClA2OA(g)+OFA2A(g) The enthalpy change is 38.55 kJ. What is the standard enthalpy of the reaction? (Δr H° = 205.6 kJ) - Chemistry

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Sum

Answer the following question.

When 6.0 g of O2 reacts with CIF as per 

\[\ce{2ClF_{(g)} + O2_{(g)} -> Cl2O_{(g)} + OF2_{(g)}}\]

The enthalpy change is 38.55 kJ. What is the standard enthalpy of the reaction? (Δr H° = 205.6 kJ)

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Solution

Given: 

Enthalpy change for a given mass = 38.55 kJ
Mass of O2 = 6.0 g

To find: Standard enthalpy of the given reaction

Calculation:

Number of moles of O2 = `("Mass of O"_2)/("Molar mass of O"_2) = (6 "g")/(32 "g mol"^-1)` = 0.1875 mol

The enthalpy change when 0.1875 mol of O2 reacts with ClF is 38.55 kJ.

∴ Enthalpy change for 1 mole O2 = `38.55/0.1875 = 205.6` kJ

From the reaction, 2 moles of ClF react with 1 mole of O2.

So, the standard enthalpy of the reaction is + 205.6 kJ.

Concept: Enthalpy (H)
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Chapter 4: Chemical Thermodynamics - Exercises [Page 89]

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Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercises | Q 4.15 | Page 89

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