Question

Calculate the enthalpy change when \(5.00 \mathrm{~g}\) of propane is burned with excess oxygen according to the reaction $$\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}\mathrm{O}(l) \quad \Delta H=-2221 \mathrm{~kJ} / \mathrm{mol}$$

Short answer

The enthalpy change for burning 5.00 g of propane with excess oxygen is \( -251.8 \mathrm{~kJ} \).

Step-by-step solution

1. Calculate moles of propane

First, we need to find the moles of propane. The molecular weight of propane (C₃H₈) is: 3 × (12.01 g/mol) for C + 8 × (1.01 g/mol) for H = 44.07 g/mol Now, we can use the given mass (5.00 g) and the molecular weight of propane to calculate the moles of propane: moles of propane = mass / molecular weight = 5.00 g / 44.07 g/mol = 0.1134 mol

2. Use stoichiometry to find the enthalpy change

We are given the standard molar enthalpy change for the reaction, which is -2221 kJ/mol. From the balanced chemical equation, we can see that 1 mole of propane reacts to produce a corresponding change in enthalpy of -2221 kJ. Now, we can use the moles of propane (0.1134 mol) and the standard molar enthalpy change to find the total enthalpy change for burning 5 g of propane: Enthalpy change = moles of propane × standard molar enthalpy change = 0.1134 mol × (-2221 kJ/mol) = -251.8 kJ

3. Report the enthalpy change

The enthalpy change for burning 5.00 g of propane with excess oxygen is -251.8 kJ.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict how much product will form from a given amount of reactants or how much of each reactant is needed to create a certain amount of product. The core of stoichiometry lies in the balanced chemical equation, where the coefficients indicate the molar ratio of the substances involved.

In the exercise provided, stoichiometry is employed to calculate the enthalpy change for the combustion of propane. The balanced equation gives a clear ratio of how many moles of propane are needed to react with oxygen, as well as how much energy is released as heat (enthalpy change) when this reaction occurs. By understanding this ratio, we can use the amount of propane we have (in moles) to find out the corresponding enthalpy change for the reaction.

Chemical Reaction

A chemical reaction involves the transformation of one set of chemical substances to another. It's characterized by the breaking and forming of bonds which leads to a change in the composition of the atoms. The equation for a chemical reaction shows the reactants (starting substances) on the left side, the products (newly formed substances) on the right side, and an arrow indicating the direction of the reaction.

The exercise describes the combustion of propane in excess oxygen, a common type of chemical reaction. Such reactions are exothermic, meaning they release heat. The balanced chemical equation not only indicates the substances involved in the chemical change but also the stoichiometric ratios and the energy change in the form of the enthalpy change.

Molar Enthalpy Change

Molar enthalpy change, often denoted as \( \Delta H \) per mole, refers to the heat released or absorbed during a chemical reaction for one mole of a substance. It is expressed in units of kilojoules per mole (kJ/mol). If the enthalpy change is negative, the reaction releases energy (exothermic); if positive, it absorbs energy (endothermic).

For the combustion of propane problem, knowing the molar enthalpy change is crucial because it allows us to calculate the total energy change when a non-standard amount of the substance is involved. The exercise clearly illustrates how to use the molar enthalpy change and the moles of the reactant to calculate the total energy change for a given quantity of fuel.