Hess’ Law and Enthalpy Cycles Revision Notes for OCR A-Level Chemistry A

Hess' Law and Enthalpy Cycles

Introduction to Hess' law

When chemists want to determine the enthalpy change of a reaction, they can sometimes measure it directly through a single experiment. However, many reactions present significant challenges for direct measurement. This is where Hess' law provides a valuable alternative approach.

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Hess' law states that when a reaction can proceed via two different routes, and both routes start and end at the same conditions, the total enthalpy change will be identical regardless of which route is taken. This principle stems from the law of conservation of energy and allows us to calculate enthalpy changes indirectly by using alternative pathways.

The power of Hess' law lies in its ability to determine unknown enthalpy changes by combining known values. If we know all enthalpy changes in a cycle except one, we can calculate the missing value using the relationship between the different routes.

Understanding enthalpy cycles

An enthalpy cycle is a diagram that visualizes the different routes a reaction can take from reactants to products. These cycles are essential tools for applying Hess' law in calculations.

In a typical enthalpy cycle, arrows represent different pathways:

Route 1 combines two steps (arrows A and B)
Route 2 shows the direct pathway (arrow C)

According to Hess' law, following the arrows from reactants to products gives us the relationship:

$A + B = C$

This means if we know any two of these enthalpy changes, we can calculate the third. This mathematical relationship is the foundation for all enthalpy cycle calculations.

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The key principle: total enthalpy change is path-independent. Whether you follow the direct route or combine multiple steps, the overall energy change remains the same.

Worked example: basic enthalpy cycle calculation

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Worked Example: Basic Enthalpy Cycle Calculation

Let's apply Hess' law to a simple enthalpy cycle to understand the calculation process.

Given information:

$A = +110 \text{ kJ mol}^{-1}$
$B = -150 \text{ kJ mol}^{-1}$
Find $C$

Step 1: Substitute the known values into the relationship $A + B = C$

Step 2: Calculate the unknown value:

C & = & A + B \\ C & = & (+110) + (-150) \\ C & = & -40 \text{ kJ mol}^{-1} \end{array}$$ **Result:** The direct route (C) has an enthalpy change of $-40 \text{ kJ mol}^{-1}$, which equals the sum of the two-step pathway.

This principle can be extended to cycles with any number of steps, provided all values except one are known.

Indirect determination using standard enthalpy of formation

Standard enthalpy changes of formation ( $\Delta_f H^\circ$ ) and combustion ( $\Delta_c H^\circ$ ) are extensively tabulated in data books. These values enable us to calculate enthalpy changes for reactions that would be difficult or impossible to measure directly.

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For elements in their standard states, $\Delta_f H^\circ$ is always zero. This is because no energy change occurs when forming an element from itself. In exams, you typically won't be provided with values of zero for elements - you're expected to know this principle.

Method: using $\Delta_f H^\circ$ values

To find the standard enthalpy change of a reaction using formation enthalpies, we construct an enthalpy cycle that connects:

The reactants
The products
Their constituent elements

The elements form the common reference point that links reactants and products together.

Worked example: reaction of iron(III) oxide with calcium

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Worked Example: Calculating Enthalpy Change Using Formation Data

Calculate the standard enthalpy change for this reaction:

$\ce{Fe2O3(s) + 3Ca(s) -> 2Fe(s) + 3CaO(s)}$

Given data:

Table

Step 1: Construct the enthalpy cycle

When building the cycle, the constituent elements $\ce{Fe(s)}$ , $\ce{Ca(s)}$ , and $\ce{O2(g)}$ form the common link. Using $\Delta_f H^\circ$ values:

The reactants and products are formed from their elements
Arrows point upwards from elements to reactants and products (formation is the reverse of breaking down)

Following the arrows from reactants to products:

Route 1: $B + A$
Route 2: $C$

By Hess' law: $B + A = C$

Therefore: $A = C - B$

Step 2: Add the $\Delta_f H$ values and calculate

A & = & (3 \times -635) - (-824) \\ A & = & -1905 + 824 \\ A & = & -1081 \text{ kJ mol}^{-1} \end{array}$$ **Result:** The negative value indicates this is an :success[exothermic reaction]. This thermite-type reaction releases substantial energy as calcium reduces iron(III) oxide.

General formula for calculations using $\Delta_f H^\circ$

For any reaction, the relationship can be expressed as:

$\Delta_r H = \sum \Delta_f H \text{ products} - \sum \Delta_f H \text{ reactants}$

The sigma ( $\sum$ ) symbol means "sum of". Remember to multiply each $\Delta_f H$ value by its stoichiometric coefficient from the balanced equation.

Indirect determination using standard enthalpy of combustion

When formation enthalpies are unavailable or impractical to use, combustion enthalpies provide an alternative route for calculations. This method is particularly useful for organic compounds.

Method: using $\Delta_c H^\circ$ values

To determine enthalpy changes using combustion data, we construct an enthalpy cycle connecting:

The reactants
The products
Their common combustion products ( $\ce{CO2(g)}$ and $\ce{H2O(l)}$ )

The combustion products form the common reference point in this type of cycle.

Worked example: formation of butane

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Worked Example: Determining Formation Enthalpy via Combustion Data

The formation of butane from its elements is shown below:

$\ce{4C(s) + 5H2(g) -> C4H10(g)}$

Direct measurement of this reaction is impossible because carbon and hydrogen form numerous different compounds simultaneously, making it impossible to isolate butane as the sole product.

However, we can measure the combustion enthalpies of $\ce{C(s)}$ , $\ce{H2(g)}$ , and $\ce{C4H10(g)}$ directly and use these to find the formation enthalpy indirectly.

Given data:

Table

Step 1: Construct the enthalpy cycle

Both the reactants and products react to form common combustion products: $\ce{CO2(g)}$ and $\ce{H2O(l)}$ . Using $\Delta_c H^\circ$ values, the arrows point downwards (combustion releases energy).

Following the arrows from reactants to products:

Route 1: $A + C$
Route 2: $B$

By Hess' law: $A + C = B$

Therefore: $A = B - C$

Step 2: Calculate using $\Delta_c H$ values

A & = & (4 \times -394) + (5 \times -286) - (-2877) \\ A & = & -1576 - 1430 + 2877 \\ A & = & -129 \text{ kJ mol}^{-1} \end{array}$$ **Result:** The formation of butane is therefore :success[exothermic], releasing $129 \text{ kJ mol}^{-1}$.

General formula for calculations using $\Delta_c H^\circ$

For any reaction, the relationship is:

$\Delta_r H = \sum \Delta_c H \text{ reactants} - \sum \Delta_c H \text{ products}$

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Notice that for combustion enthalpies, we subtract products from reactants (the opposite order to formation enthalpies). This is because the arrows in combustion cycles point downwards, whereas in formation cycles they point upwards.

Summary of calculation methods

Two key formulas enable you to calculate enthalpy changes indirectly:

Using enthalpy changes of formation:

$\Delta_r H = \sum \Delta_f H \text{ products} - \sum \Delta_f H \text{ reactants}$

Using enthalpy changes of combustion:

$\Delta_r H = \sum \Delta_c H \text{ reactants} - \sum \Delta_c H \text{ products}$

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The order is reversed between these two methods due to the direction of arrows in their respective enthalpy cycles. In formation cycles, arrows point upward from elements. In combustion cycles, arrows point downward to combustion products.

Working with unfamiliar enthalpy cycles

Exam questions may present complex enthalpy cycles involving multiple steps and unfamiliar reactions. The key to solving these is systematic analysis.

General approach:

Identify all the steps in the cycle and label the enthalpy changes
Match each enthalpy change to values provided in data tables
Determine the two possible routes through the cycle
Apply Hess' law: the sum of enthalpy changes for each route must be equal
Rearrange to solve for the unknown enthalpy change

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Even when cycles appear complicated with multiple pathways, the fundamental principle remains: total enthalpy change is independent of route taken. Focus on identifying which values are known, trace the possible routes carefully, and set up your equation accordingly.

Exam tips

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Common mistakes to avoid:

Forgetting that $\Delta_f H^\circ = 0$ for elements in their standard state
Using the wrong sign when rearranging equations
Confusing the formulas for formation and combustion calculations
Forgetting to multiply by stoichiometric coefficients

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Top tips:

Always draw the enthalpy cycle clearly, even if one is provided
Label your arrows and follow them systematically
Check your final answer makes chemical sense (is it reasonable?)
Remember: formation arrows point UP, combustion arrows point DOWN
Keep track of positive and negative signs throughout

Remember!

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Key Points to Remember:

Hess' law states that total enthalpy change is independent of the route taken, allowing indirect determination of enthalpy changes that cannot be measured directly.
Enthalpy cycles are diagrams showing alternative routes between reactants and products. If all values except one are known, the unknown can always be calculated.
Using formation enthalpies: $\Delta_r H = \sum \Delta_f H \text{ (products)} - \sum \Delta_f H \text{ (reactants)}$ . Formation cycles connect via elements, with arrows pointing upward.
Using combustion enthalpies: $\Delta_r H = \sum \Delta_c H \text{ (reactants)} - \sum \Delta_c H \text{ (products)}$ . Combustion cycles connect via combustion products, with arrows pointing downward.
Always multiply tabulated values by their stoichiometric coefficients and remember that $\Delta_f H^\circ = 0$ for all elements in their standard states.

Hess’ Law and Enthalpy Cycles (OCR A-Level Chemistry A): Revision Notes