Question

sulfur reacts with oxygen to form sulfur dioxide (so₂(g), δhf = -296.8 kj/mol) according to the equation below.
s(s) + o₂(g) → so₂(g)
what is the enthalpy change for the reaction?
use δhᵣₓₙ = σ(δhf,products) - σ(δhf,reactants).
○ -593.6 kj
○ -296.8 kj
○ 296.8 kj
○ 593.6 kj

Explanation:

Step1: Identify $\Delta H_f$ values

For reactants: $S(s)$ is an element in its standard state, so $\Delta H_{f,S(s)} = 0\ \text{kJ/mol}$; $O_2(g)$ is also an element in its standard state, so $\Delta H_{f,O_2(g)} = 0\ \text{kJ/mol}$.
For product: $\Delta H_{f,SO_2(g)} = -296.8\ \text{kJ/mol}$.

Step2: Apply the formula $\Delta H_{rxn} = \sum(\Delta H_{f,products}) - \sum(\Delta H_{f,reactants})$

Calculate $\sum(\Delta H_{f,products})$: $1 \times (-296.8) = -296.8\ \text{kJ/mol}$.
Calculate $\sum(\Delta H_{f,reactants})$: $1 \times 0 + 1 \times 0 = 0\ \text{kJ/mol}$.
Substitute into the formula: $\Delta H_{rxn} = -296.8 - 0 = -296.8\ \text{kJ/mol}$.

Answer:

-296.8 kJ (corresponding to the option: -296.8 kJ)