IB DP Chemistry: HL复习笔记5.2.3 Using Hess's Law - Equations
Hess's Law using Equations
- We can use Hess’s Law to solve unknown enthalpy changes by combining equations
- This requires a methodical step-by-step approach
- It is necessary to identify how the given equations relate to the target equation as the following example illustrates
Solving Hess's Law problems using equations step-by-step :
Worked Example
Consider the following reactions.
N2(g) + O2(g) → 2NO(g) ∆H = +180 kJ
2NO2(g) → 2NO(g) + O2(g) ∆H = +112 kJ
What is the ∆H value, in kJ, for the following reaction?
N2(g) + 2O2(g) → 2NO2(g)
Answer:
1. Identify which given equation contains the product you want
This equation contains the desired product on the left side:
2NO2(g) → 2NO(g) + O2(g) ∆H = +112 kJ
2. Adjust the equation if necessary, to give the same product. If you reverse it, reverse the ΔH value
Reverse it and reverse the sign
2NO(g) + O2(g) → 2NO2(g) ∆H = -112 kJ
3. Adjust the equation if necessary, to give the same number of moles of product
The equation contains the same number of moles as in the question, so no need to adjust the moles
Next steps
4. Identify which given equation contains your reactant
This equation contains the reactant
N2(g) + O2(g) → 2NO(g) ∆H = +180 kJ
5. Adjust the equation if necessary, to give the same reactant. If you reverse it, reverse the ΔH value
No need to reverse it as the reactant is already on the left side
6. Adjust the equation if necessary, to give the same number of moles of reactant
Final steps
7. Add the two equations together
N2(g) + O2(g) → 2NO(g) ∆H = +180 kJ
2NO(g) + O2(g) → 2NO2(g) ∆H = -112 kJ
8. Cancel the common items
N2(g) + O2(g) + 2NO(g) + O2(g) → 2NO(g) + 2NO2(g)
9. Add the two ΔH values together to get the one you want
N2(g) + 2O2(g) → 2NO2(g) ∆H = +180-112 = +68 kJ
Worked Example
The enthalpy changes for two reactions are given.
Br2 (l) + F2 (g) → 2BrF (g) ΔH = x kJ
Br2 (l) + 3F2 (g) → 2BrF3 (g) ΔH = y kJWhat is the enthalpy change for the following reaction?
BrF (g) + F2 (g) → BrF3 (g)
A. x – y
B. y - x
C. ½ (–x + y)
D. ½ (x – y)
Answer:
The correct option is C.
-
- The second equation contains the desired product, but it needs to be halved to make 1 mole
Br2 (l) + 3F2 (g) → 2BrF3 (g) ΔH = y becomes
½Br2 (l) + 1½F2 (g) → BrF3 (g) ½ΔH = ½y
-
- The first equation contains the reactant, but it needs to be reversed and halved:
Br2 (l) + F2 (g) → 2BrF (g) ΔH = x becomes
BrF (g) → ½Br2 (l) + ½F2 (g) ½ΔH = -½x
-
- Combine the two equations and cancel the common terms:
½Br2 (l) + 1½F2 (g) → BrF3 (g) ½ ΔH = y kJ
BrF (g) → ½Br2 (l) + ½F2 (g) ½ ΔH = -x kJ
BrF (g) + F2 (g) → BrF3 (g) ΔH = ½y + -½x = ½(-x + y)
Exam Tip
If doesn't matter whether you use equations or cycles to solve Hess's Law problems, but you should be familiar with both methods and sometimes one is easier than another
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