Answer:
∆H for the reaction is -108.7 KJ
Step-by-step explanation:
To calculate ∆H of a reaction using Hess’ law , we subtract the ∆H of reactants from that of the products
What we should however understand is that the ∆H of elements standing alone and the ∆H of molecules of gaseous elements equals zero.
So let’s proceed
For the question, what we want to calculate is
∆H = ∆H(ClF3) - ∆H(ClF)
Kindly note that ∆H(F2) = 0, that’s why I am not including it in the calculation
We can shop for the ∆H of both the reactants and the products from the reactions below;
For the first one, the expression is
∆H = ∆H(F20) + ∆H(Cl20) - 2∆H(ClF) = 167.4
For the second, we have
∆H = 3∆H(F20) + ∆H(Cl2O) - 2∆H(ClF3) = 341.4
and lastly, we have;
∆H = 2∆H(F2O) = -43.4
So how do we proceed from here? To cancel out the ∆H(Cl20) and ∆H(F20), we add the 1st and last while we subtract the second.
So we have ;
∆H(F20) + ∆H(Cl20) - 2∆H(ClF) + 2∆H(F2O) - {3∆H(F20) + ∆H(Cl2O) - 2∆H(ClF3)} = (167.4 -43.4)-341.4
So we have ∆H(Cl20) and ∆H(F20) canceled out. we are left with;
-2∆H(ClF) + 2∆H(ClF3) = -217.4
Divide through by 2, we have;
∆H(ClF3) - ∆H(ClF) = -108.7 KJ
Now compare this with the initial equation we first wrote and you can see that they are equal