IB DP Chemistry: SL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)         ∆= +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)


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)         ∆= +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)                                             ∆= +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)                         ∆ = +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)


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