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Question
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.
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