Given the combustion reaction below (combustion reactions are when a substance reacts with oxygen):

Question

2H₂(g) + O₂(g) ---> 2H₂O(ℓ) ΔH = -570 kJ

1. H_2(g) + ½ O_2(g) → H₂O_(ℓ)  ΔH_comb = −285 kJ/mol.
2. 2H_2(g) + O_2(g) → 2H₂O_(ℓ)   ΔH_comb = −285 kJ/mol.
3. 2H_2(g) + O_2(g) → 2H₂O_(ℓ)   ΔH_comb = −570 kJ/mol.
4. H_2(g) + ½ O_2(g) → H₂O_(ℓ)  ΔH_comb = −570 kJ/mol.

Answer 1

This is an application of Hess's Law. The heat of reaction, ΔH, scales with the coefficients in the balanced chemical equation. If you multiply the entire equation by a factor, then ΔH must also be multiplied by the same factor. For example, if the coefficients for all the reactants and products are multiplied by 2 compared to the original balanced reaction, then the ΔH must also be multiplied by 2. Let’s discuss each option and evaluate its accuracy.

1.     Correct. This is a balanced reaction where all the coefficients are exactly half of the original reaction and ΔH is correctly calculated as half of the original reaction’s ΔH. ΔH=-570kJ/2=-285kJ.

2.     Incorrect. This is the same balanced equation as the original, but ΔH changed. ΔH should remain -570kJ.

3.     Correct. This is the same as the original reaction.

4.     Incorrect. This reaction is correctly balanced and all the coefficients are exactly half of the original reaction, but ΔH needed to be half as much as well.

You can think of scaling ΔH like you would think of scaling a recipe for cooking. If you change how much “recipe” you make, you will change how much heat is involved. If you double the ingredients in a batch, you double the output. Similarly, if you double all the reactants in a combustion reaction and you double all the products, you can expect double the heat output.