Use Hess’s Law to determine the enthalpy change (∆H) for the reaction: ClF + F2 → ClF3 Given:
2ClF + O2 → Cl2O + F2O. ∆H=167.4kJ
2ClF3 + 2O2 →Cl2O + 3F2O. ∆H=341.4kJ
2F2 + O2 →2F2O. ∆H=-43.4kJ

∆H for the reaction is -108.7 KJ

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

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer