Using the bond energy data from your text (or the internet), determine (show calculations for) the approximate enthalpy change , ∆H, for each of the following reactions: (a) Cl2 (g) + 3F2 (g) ⟶ 2ClF3 (g)

∆H=  438 KJ/mol

Explanation:

First, we have to find the energy bond values for each compound:

-) Cl-Cl = 243 KJ/mol

-) F-F = 159 KJ/mol

-) F-Cl = 193 KJ/mol

If we check the reaction we can calculate the number of bonds:

In total we will have:

-) Cl-Cl = 1

-) F-F = 3

-) F-Cl = 6

With this in mind. we can calculate the total energy for each bond:

-) Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol

-) F-F = (3*159 KJ/mol) = 477 KJ/mol

-) F-Cl = (6*193 KJ/mol) = 1158 KJ/mol

Now, we can calculate the total energy of the products and the reagents:

Reagents = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol

Products = 1158 KJ/mol

Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:

∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol

I hope it helps!

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