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Solution: Reaction of gaseous ClF with F 2 yields liquid ClF3, an important fluorinating agent. Use the following thermochemical equations to calculate ΔH°rxn for this reaction:(1) 2CIF(g) + O2(g) ⟶ Cl2O(g) + OF2(g)          ΔH°rxn = 167.5 kJ(2) 2F2(g) + O2(g) ⟶ 2OF2(g)                         ΔH°rxn = −43.5 kJ(3) 2ClF3(l) + 2O2(g) ⟶ Cl2O(g) + 3OF2(g)    ΔH°rxn = 394.1 kJ2ClF  + 2F2 ⟶  2ClF3

Problem

Reaction of gaseous ClF with F 2 yields liquid ClF3, an important fluorinating agent. Use the following thermochemical equations to calculate ΔH°rxn for this reaction:

(1) 2CIF(g) + O2(g) ⟶ Cl2O(g) + OF2(g)          ΔH°rxn = 167.5 kJ
(2) 2F2(g) + O2(g) ⟶ 2OF2(g)                         ΔH°rxn = −43.5 kJ
(3) 2ClF3(l) + 2O2(g) ⟶ Cl2O(g) + 3OF2(g)    ΔH°rxn = 394.1 kJ

2ClF  + 2F2 ⟶  2ClF3

Solution

To find the ΔHrxn, we need to use the Hess's Law where we have to rearrange the given equations to get the desired reaction.

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