Answer:
∆H= <u>438 KJ/mol</u>
Explanation:
First, we have to find the <u>energy bond values</u> 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 <u>number of bonds</u>:
In total we will have:
-) Cl-Cl = 1
-) F-F = 3
-) F-Cl = 6
With this in mind. we can calculate the <u>total energy for each bond</u>:
-) 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 <u>products</u> and the <u>reagents</u>:
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 <u>subtraction</u> between products and reagents:
∆H= 1158 KJ/mol-720 KJ/mol = <u>438 KJ/mol</u>
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I hope it helps!