Respuesta :
Answer:
∆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:
[tex]Cl_2_(_g_)~+~3F_2_(_g_)~->~2ClF_3_(_g_)[/tex]
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!
The approximate enthalpy change is:
∆H= 438 KJ/mol
Calculation for enthalpy change:
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
Balanced chemical reaction:
Cl₂ (g) + 3F₂ (g) ⟶ 2ClF₃ (g)
Total number of bond for each:
Cl-Cl = 1
F-F = 3
F-Cl = 6
Total bond energy will be:
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 reactants:
Reactants = 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
Find more information about enthalpy change here:
brainly.com/question/14047927
