David V. answered • 03/03/21

PhD in ChemE with 9+ Years of Research and Consulting Experience

It would be helpful to know what your general approach to this problem was, but let's walk through the suggested steps. I think you were on the right track, but maybe there was a typo along the way. To find the standard heat of reaction for the formation of ClF_{3} by fluorinating ClF, you need to rearrange the three provided individual reactions to give the desired reaction. Three general guidelines:

-To start off, try adding the three provided reactions as is.

-If needed, you can multiply the reactant/product coefficients by an integer to get compounds to cancel out. In this case, also multiply the ΔH^{0} by the same integer.

-If needed, you can reverse any individual reaction. In this case, flip the sign of the ΔH^{0}.

Looking at the three reactions, I see that the third reaction should be flipped because it is the only instance of ClF_{3}, but it is a reactant instead of a product. Its ΔH^{0} should now be -394.1 kJ instead of +394.1 kJ. When we add the three reactions together, we get:

2 CIF(g) + O_{2}(g) ---> Cl_{2}O(g) + OF_{2}(g) ΔH^{0} = 167. 5 kJ

2 F_{2}(g) + O_{2}(g) ---> 2 OF_{2}(g) ΔH^{0} = -43.5 kJ

Cl_{2}O(g) + 3 OF_{2}(g) ---> 2 CIF_{3}(l) + 2 O_{2}(g) ΔH^{0} = -394.1 kJ

2 CIF(g) + 2 F_{2}(g) ---> 2 CIF_{3}(l) ΔH^{0} = 167.5 - 43.5 - 394.1 = -270.6 kJ

You'll also notice that all three compounds in the final reaction of a coefficient of 2. This can be simplified by dividing all three coefficients and the ΔH^{0} by 2. So the ΔH^{0} for the production of ClF3 is -135.3 kJ/mole.