# IB DP Chemistry: SL复习笔记5.2.2 Using Hess's Law - Cycles

### Hess's Law using Cycles

• There are two common methods to solving Hess's Law problems, using cycles and using equations
• To be successful in using cycles you need to follow carefully a step-by-step plan using the information in the question to construct a cycle and add the given information
• The following example shows one way to lay out your solution:

#### Worked Example

Calculate the enthalpy of reaction for

2N2(g) + 6H2(g)  4NH3(g)

Given the data:4NH3(g) + 3O2(g)  2N2(g) + 6H2O(l),           ΔH1 = -1530 kJ mol-1

H2(g) + ½O2(g )  H2O(l),                               ΔH2 = -288 kJ mol-1

• Begin by writing the target enthalpy change at the top of your diagram from left to right:

• Next, write the alternative route at the bottom of your cycle and connect the top and bottom with arrows pointing in the correct directions:

• Add the enthalpy data and adjust, as necessary, for different molar amounts

• Write the Hess's Law calculation out:

ΔH= +6ΔH2 – ΔH= + (-288 x 6) - ( -1530 ) = -198 kJ

• Two important rules:
• If you follow the direction of the arrow you ADD the quantity
• If you go against the arrow you SUBTRACT the quantity

#### Worked Example

What is the enthalpy change, in kJ, for the reaction below?

4FeO(s) + O2(g) → 2Fe2O3(s)

Given the data:2Fe(s) + O2(g) → 2FeO(s)                  ∆H = – 544 kJ

4Fe(s) + 3O2(g) → 2Fe2O3(s)             ∆H = –1648 kJ

• Draw the Hess cycle and add the known values

• Write the Hess's Law calculation out:

Follow the alternative route and the process the calculation

ΔH= - ( - 544 x 2) + (- 1648) = - 560 kJ

#### Exam Tip

It is very important you get the arrows in the right direction and that you separate the mathematical operation from the sign of the enthalpy change. Many students get these problems wrong because they confuse the signs with the operations. To avoid this always put brackets around the values and add the mathematical operator in front