Law of Conservation of Energy Revision Notes for HSC SSCE Chemistry

Law of conservation of energy

What is the law of conservation of energy?

The law of conservation of energy is a fundamental principle in science stating that energy cannot be created or destroyed. However, energy can be transformed from one form to another. For example, chemical energy can be converted to heat energy, or electrical energy can be changed to light energy.

In chemistry, this law has an important application. When a chemical reaction occurs, the total energy change between reactants and products remains constant, no matter which pathway the reaction follows. This chemical version of the law of conservation of energy is called Hess's law, named after the chemist Germain Hess who first recognised this principle.

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Understanding the conservation of energy is essential for grasping why Hess's law works. Since energy cannot appear or disappear, the total energy change in any process must be the same regardless of how many steps are involved or what route is taken.

Understanding Hess's law

Hess's law states that the enthalpy change ( $\Delta H$ ) for a chemical reaction is the same regardless of the number of steps taken to complete the reaction. Whether you go from reactants to products in one step or through multiple intermediate steps, the total energy change will be identical.

This is incredibly useful because it allows us to calculate enthalpy changes for reactions that are difficult or impossible to measure directly. We can use known enthalpy changes from other reactions to work out the unknown value.

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Key Principle of Hess's Law:

The total enthalpy change is independent of the reaction pathway. This means you can:

Break a reaction into multiple steps
Add the enthalpy changes of those steps
Get the same result as the direct pathway

This principle is a direct consequence of the conservation of energy!

Example: carbon combustion

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Worked Example: Applying Hess's Law to Carbon Combustion

Let's explore Hess's law using the combustion of carbon as an example.

Indirect pathway (two steps):

When carbon burns in a limited supply of oxygen, carbon monoxide forms first:

Step 1: $2\text{C}(\text{s}) + \text{O}_2(\text{g}) \rightarrow 2\text{CO}(\text{g})$ where $\Delta H_1 = -222 \text{ kJ mol}^{-1}$

This carbon monoxide can then react with more oxygen to form carbon dioxide:

Step 2: $2\text{CO}(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g})$ where $\Delta H_2 = -564 \text{ kJ mol}^{-1}$

When we add these two reactions together, the $2\text{CO}(\text{g})$ appears on both sides and cancels out. Collecting the oxygen molecules gives us:

Overall reaction: $2\text{C}(\text{s}) + 2\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g})$

The total enthalpy change for this indirect pathway is:

$\Delta H_3 = \Delta H_1 + \Delta H_2 = -222 + (-564) = -786 \text{ kJ mol}^{-1}$

Direct pathway (one step):

Alternatively, carbon can burn directly in excess oxygen to form carbon dioxide:

$\text{C}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g})$ where $\Delta H_4 = -393 \text{ kJ mol}^{-1}$

This value is per mole of carbon dioxide. To compare it with the indirect route (which produced two moles of carbon dioxide), we need to double it:

$2\text{C}(\text{s}) + 2\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g})$ where $\Delta H_5 = -2 \times 393 = -786 \text{ kJ mol}^{-1}$

Conclusion: Both pathways give the same enthalpy change of -786 kJ per two moles of CO₂. This demonstrates Hess's law perfectly!

Drawing and using enthalpy diagrams

Enthalpy diagrams are visual tools that help us analyse energy changes in chemical reactions. They are extremely useful for understanding Hess's law problems.

How to draw an enthalpy diagram:

Draw horizontal lines to represent different energy states:
- Initial state (reactants)
- Intermediate state(s) (if any)
- Final state (products)
Label each line with the chemical substances:
- Place all reactants on the initial state line
- Place all products on the final state line
- Place intermediate substances on their respective lines
Draw arrows between states:
- Arrows show the direction of change
- Point from one energy level to another
Label the arrows with enthalpy changes ( $\Delta H$ values)

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Reading Enthalpy Diagrams:

The enthalpy change shown on each arrow represents the increase in enthalpy when moving from the tail of the arrow to the head of the arrow. When following a series of arrows in sequence, the total enthalpy change is the sum of all individual changes, regardless of whether the arrows point up or down the page.

For example, in the carbon combustion diagram above, you can follow two different paths from $2\text{C}(\text{s}) + 2\text{O}_2(\text{g})$ to $2\text{CO}_2(\text{g})$ :

Direct path: One arrow showing $-786 \text{ kJ mol}^{-1}$
Indirect path: Two arrows showing $-222 \text{ kJ mol}^{-1}$ then $-564 \text{ kJ mol}^{-1}$

Both paths give the same total energy change.

Rules for manipulating equations

When using Hess's law to calculate unknown enthalpy changes, you need to manipulate chemical equations. Two important rules apply:

Rule 1: Reversing a reaction changes the sign of ΔH

If you write a chemical equation in the opposite direction, you must change the sign of the enthalpy change.

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Reversing Reactions:

When you reverse a chemical equation, the sign of $\Delta H$ must flip:

Forward: $\text{A} \rightarrow \text{B}$ where $\Delta H = -50 \text{ kJ mol}^{-1}$
Reverse: $\text{B} \rightarrow \text{A}$ where $\Delta H = +50 \text{ kJ mol}^{-1}$

This makes sense because if energy is released going forward (negative $\Delta H$ ), then energy must be absorbed going backward (positive $\Delta H$ ).

Rule 2: Multiplying or dividing a reaction multiplies or divides ΔH

If you double the coefficients in a chemical equation, you must double the enthalpy change. If you halve the coefficients, you must halve the enthalpy change.

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Scaling Reactions:

When you multiply or divide the coefficients in an equation, you must do the same to $\Delta H$ :

Original: $\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l})$ where $\Delta H = -286 \text{ kJ mol}^{-1}$
Doubled: $2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{l})$ where $\Delta H = -572 \text{ kJ mol}^{-1}$

The enthalpy change scales proportionally with the amount of substance reacting.

Worked example: phosphorus chlorides

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Worked Example: Calculating Unknown Enthalpy Changes

Problem: When excess phosphorus reacts with a small amount of chlorine, gaseous phosphorus trichloride forms with an enthalpy change of $-287 \text{ kJ mol}^{-1}$ . When a small amount of phosphorus reacts with excess chlorine, gaseous phosphorus pentachloride forms with an enthalpy change of $-375 \text{ kJ mol}^{-1}$ . Calculate the enthalpy change for the reaction:

$\text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow \text{PCl}_5(\text{g})$

Solution:

Step 1: Write down the given information

Equation 1: $\text{P}(\text{s}) + 1\frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{PCl}_3(\text{g})$ where $\Delta H_1 = -287 \text{ kJ mol}^{-1}$

Equation 2: $\text{P}(\text{s}) + 2\frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{PCl}_5(\text{g})$ where $\Delta H_2 = -375 \text{ kJ mol}^{-1}$

Step 2: Write the target equation

Equation 3: $\text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow \text{PCl}_5(\text{g})$ where $\Delta H_3 = ?$

Step 3: Plan the manipulation

To create equation 3, we need $\text{PCl}_3$ on the left side and $\text{PCl}_5$ on the right side. Currently, in equation 1, $\text{PCl}_3$ is on the right side, so we must reverse equation 1.

Step 4: Reverse equation 1

Equation 4: $\text{PCl}_3(\text{g}) \rightarrow \text{P}(\text{s}) + 1\frac{1}{2}\text{Cl}_2(\text{g})$ where $\Delta H_4 = +287 \text{ kJ mol}^{-1}$

Note that the sign has changed from negative to positive.

Step 5: Add equations 4 and 2

$\text{PCl}_3(\text{g}) + \text{P}(\text{s}) + 2\frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{P}(\text{s}) + 1\frac{1}{2}\text{Cl}_2(\text{g}) + \text{PCl}_5(\text{g})$

Step 6: Cancel common terms

$\text{P}(\text{s})$ appears on both sides, so it cancels out.

$1\frac{1}{2}\text{Cl}_2(\text{g})$ cancels from the $2\frac{1}{2}\text{Cl}_2(\text{g})$ , leaving $\text{Cl}_2(\text{g})$ on the left side.

This gives us: $\text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow \text{PCl}_5(\text{g})$

Step 7: Calculate the enthalpy change

$\Delta H_3 = \Delta H_4 + \Delta H_2 = +287 + (-375) = -88 \text{ kJ mol}^{-1}$

Verification using enthalpy diagram:

From the enthalpy diagram, we can see that:

$-287 + \Delta H_3 = -375$

Therefore:

$\Delta H_3 = -375 - (-287) = -375 + 287 = -88 \text{ kJ mol}^{-1}$

This confirms our calculation - the answer is -88 kJ mol⁻¹

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About Fractional Coefficients:

In thermochemistry, we often write equations with fractions (like $1\frac{1}{2}\text{Cl}_2$ ). This is acceptable because we read these equations in terms of moles, and we can have half a mole of a substance. Remember, we cannot have half a molecule, but we can discuss half a mole of molecules.

Exam tips

Always check whether you need to reverse, multiply, or divide equations to match your target equation
Remember to change the sign of ΔH when you reverse a reaction
Draw an enthalpy diagram if you find the problem confusing—visual representation often makes the solution clearer
Check that substances appearing on both sides of your combined equation cancel out properly
Pay attention to coefficients and make sure you scale $\Delta H$ values appropriately
When given enthalpy per mole, check what "per mole" refers to (per mole of which substance)

Remember!

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Key Points to Remember:

Energy conservation: Energy cannot be created or destroyed, only transformed from one form to another
Hess's law: The total enthalpy change between reactants and products is the same regardless of the pathway taken
Reversing reactions: If you reverse a reaction, change the sign of $\Delta H$ (negative becomes positive, positive becomes negative)
Scaling reactions: If you multiply or divide a reaction equation by a number, multiply or divide $\Delta H$ by the same number
Enthalpy diagrams: These are powerful visual tools that show all possible pathways between reactants and products, making it easier to apply Hess's law

Law of Conservation of Energy (HSC SSCE Chemistry): Revision Notes