What is Hess law — calculate delta H using two given reactions

Question

State Hess’s Law. Using the following two reactions, calculate $\Delta H$ for the reaction: $C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$

Given:

$C(s) + O_2(g) \rightarrow CO_2(g)$ ; $\Delta H_1 = -393.5$ kJ/mol
$CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$ ; $\Delta H_2 = -283.5$ kJ/mol

Solution — Step by Step

Hess’s Law of Constant Heat Summation states: The total enthalpy change for a reaction is the same, regardless of whether the reaction takes place in one step or multiple steps — as long as the initial and final states are the same.

This is a consequence of enthalpy being a state function — it depends only on the initial and final states, not on the path taken between them.

We want: $C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$ … call this $\Delta H = ?$

We have reactions 1 and 2. We need to manipulate them to arrive at the target reaction.

Keep Reaction 1 as is:

C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_1 = -393.5 \text{ kJ/mol}

Reverse Reaction 2 (because we need CO on the product side, not reactant side; reversing flips the sign of $\Delta H$ ):

CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g) \quad \Delta H_2' = +283.5 \text{ kJ/mol}

Add Reaction 1 and Reversed Reaction 2:

C(s) + O_2(g) + CO_2(g) \rightarrow CO_2(g) + CO(g) + \frac{1}{2}O_2(g)

Cancel species that appear on both sides ( $CO_2$ cancels, half mole of $O_2$ cancels):

C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g) \checkmark

This is exactly our target reaction.

\Delta H = \Delta H_1 + \Delta H_2' = -393.5 + (+283.5) = -110.0 \text{ kJ/mol}

Answer: $\Delta H = -110.0$ kJ/mol

The formation of CO from carbon is exothermic, releasing 110.0 kJ per mole.

Why This Works

Enthalpy is a state function — like altitude on a map. Whether you travel from Mumbai to Delhi via Pune or via Jaipur, the altitude difference (initial minus final) is the same. Similarly, whether CO forms directly from C, or via a detour through $CO_2$ , the enthalpy difference is the same.

Hess’s Law allows us to calculate $\Delta H$ for reactions that are experimentally difficult or dangerous to perform directly. We combine reactions we CAN measure to get the enthalpy of a reaction we can’t.

The two operations you’re allowed with Hess’s Law: (1) reverse a reaction → change sign of $\Delta H$ ; (2) multiply a reaction by a factor → multiply $\Delta H$ by same factor. These two operations, applied to available reactions, should give you the target reaction when added together.

Alternative Method — Using Formation Enthalpies

If standard enthalpies of formation are given:

\Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})

For our target reaction: $C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$

$\Delta H_f[C(s)] = 0$ (element in standard state) $\Delta H_f[O_2(g)] = 0$ (element in standard state) $\Delta H_f[CO(g)] = -110.5$ kJ/mol (standard value)

\Delta H = -110.5 - 0 - 0 = -110.5 \text{ kJ/mol}

Consistent with our Hess’s Law calculation (small difference due to rounding of given values).

Common Mistake

When reversing a reaction, students sometimes forget to flip the sign of $\Delta H$ , or they flip it but then add incorrectly. Remember: if you reverse a reaction, both the equation AND the sign of $\Delta H$ change. A reaction that is exothermic in the forward direction is endothermic in the reverse direction. Forgetting this sign flip is the most common error in Hess’s Law problems — check sign changes explicitly after each manipulation.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Formation Enthalpies

Common Mistake

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