Example 5.14: A More Challenging Problem Using Hess’s Law
Chlorine monofluoride can react with fluorine to form chlorine trifluoride:
[1]
\(\ce{ClF(g)}\)\(\ce{ + }\)\(\ce{F2(g)}\)\(\ce{->}\)\(\ce{ClF3(g)}\)\(\ce{ }\) \(ΔH_{\mathrm{1}} = \)?
Use the reactions here to determine the
\(Δ\)H for reaction [1]:
[2]
\(\ce{2OF2(g)}\)\(\ce{->}\)\(\ce{O2(g)}\)\(\ce{ + }\)\(\ce{2F2(g)}\)\(\ce{ }\) \(ΔH_{\mathrm{2}} = - 49.4\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
[3]
\(\ce{2ClF(g)}\)\(\ce{ + }\)\(\ce{O2(g)}\)\(\ce{->}\)\(\ce{Cl2O(g)}\)\(\ce{ + }\)\(\ce{OF2(g)}\)\(\ce{ }\) \(ΔH_{\mathrm{3}} = 205.6\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
[4]
\(\ce{ClF3(g)}\)\(\ce{ + }\)\(\ce{O2(g)}\)\(\ce{->}\)\(\ce{1/2Cl2O(g)}\)\(\ce{ + }\)\(\ce{3/2OF2(g)}\)\(\ce{ }\) \(ΔH_{\mathrm{4}} = 266.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
Solution
Our goal is to manipulate and combine reactions [2], [3], and [4] such that they add up to reaction [1]. Going from left to right in [1], we first see that
\(\ce{ClF(g)}\) is needed as a reactant. This can be obtained by multiplying reaction [3] by 1/2, which means that the
\(ΔH\) change is also multiplied by 1/2:
\(ΔH_{\mathrm{3}}\) \(= 205.6\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(ΔH_{\mathrm{3a}}\) \(= \frac{1 }{ 2} \cdot ΔH_{\mathrm{3}}\)
\(\ \ \ =\frac{1 }{ 2} \cdot 205.6\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ \ \ =102.80\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
[3a]
\(\ce{ClF(g)}\)\(\ce{ + }\)\(\ce{1/2O2(g)}\)\(\ce{->}\)\(\ce{1/2Cl2O(g)}\)\(\ce{ + }\)\(\ce{1/2OF2(g)}\)\(\ce{ }\) \(ΔH\) \( = 102.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
Next, we see that
\(\ce{F2}\) is also needed as a reactant. To get this, reverse and halve reaction [2], which means that the
\(ΔH\) changes sign and is halved:
\(ΔH_{\mathrm{2}}\) \(= -49.4\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(ΔH_{\mathrm{2a}}\) \(= -\frac{1 }{ 2} \cdot ΔH_{\mathrm{2}}\)
\(\ \ \ =-\frac{1 }{ 2} \cdot (-49.4\ \frac{\mathrm{kJ}}{\mathrm{mol}})\)
\(\ \ \ =24.70\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
[2a]
\(\ce{1/2O2(g)}\)\(\ce{ + }\)\(\ce{F2(g)}\)\(\ce{->}\)\(\ce{OF2(g)}\)\(\ce{ }\) \(ΔH\)\( = 24.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\).
To get
\(\ce{ClF3}\) as a product, reverse [4], changing the sign of
\(ΔH:\)
\(ΔH_{\mathrm{4}}\) \(= 266.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(ΔH_{\mathrm{4a}}\) \(= -ΔH_{\mathrm{4}}\)
\(\ \ \ =-266.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
[4a]
\(\ce{1/2Cl2O(g)}\)\(\ce{ + }\)\(\ce{3/2OF2(g)}\)\(\ce{->}\)\(\ce{ClF3(g)}\)\(\ce{ + }\)\(\ce{O2(g)}\)\(\ce{ }\) \(ΔH\)\( = - 266.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
Now check to make sure that these reactions add up to the reaction we want. Without cancelling, we have (the equation is balanced because we added up balanced equations):
[3a+2a+4a]
\(\ce{ClF}\)\(\ce{ + }\)\(\ce{1/2O2}\)\(\ce{ + }\)\(\ce{1/2O2}\)\(\ce{ + }\)\(\ce{F2}\)\(\ce{ + }\)\(\ce{1/2Cl2O}\)\(\ce{ + }\)\(\ce{3/2OF2}\)\(\ce{->}\)\(\ce{1/2Cl2O}\)\(\ce{ + }\)\(\ce{1/2OF2}\)\(\ce{ + }\)\(\ce{OF2}\)\(\ce{ + }\)\(\ce{ClF3}\)\(\ce{ + }\)\(\ce{O2}\)\(\ce{ }\)
Reactants 1/2
\(\ce{O2}\) and 1/2
\(\ce{O2}\) cancel out product O2; product 1/2
\(\ce{Cl2O}\) cancels reactant 1/2
\(\ce{Cl2O}\) ; and reactant 3/2
\(\ce{OF2}\) is cancelled by products 1/2
\(\ce{OF2}\) and
\(\ce{OF2}\). This leaves only reactants
\(\ce{ClF}\) and
\(\ce{F2}\) and product
\(\ce{ClF3}\), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified
\(ΔH\) values will give the desired
\(ΔH:\)
\(ΔH\) \(= ΔH_{\mathrm{3a}} + ΔH_{\mathrm{2a}} + ΔH_{\mathrm{4a}}\)
\(\ \ \ =102.80\ \frac{\mathrm{kJ}}{\mathrm{mol}} + 24.70\ \frac{\mathrm{kJ}}{\mathrm{mol}} + -266.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ \ \ =127.500\ \frac{\mathrm{kJ}}{\mathrm{mol}} + -266.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(\ \ \ =-139.2\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)