Question

Consider the following reaction: $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{kJ}$$ a. How much heat is evolved for the production of 1.00 mole of \(\mathrm{H}_{2} \mathrm{O}(l) ?\) b. How much heat is evolved when 4.03 g hydrogen are reacted with excess oxygen? c. How much heat is evolved when \(186 \mathrm{g}\) oxygen are reacted with excess hydrogen?

Short answer

a. The heat evolved for the production of 1.00 mole of H2O is -286 kJ/mol. b. The heat evolved when 4.03 g of hydrogen is reacted with excess oxygen is -576.29 kJ. c. The heat evolved when 186 g of oxygen is reacted with excess hydrogen is -3326.5 kJ.

Step-by-step solution

a. Heat evolved for 1 mole H2O

We are given a balanced equation and the enthalpy change \(\Delta H\). To find the heat evolved for the production of 1 mole of H2O, we need to find the heat evolved for the stoichiometric production of the reaction and then divide it by the stoichiometric coefficient of water. According to the balanced equation, the reaction produces 2 moles of water, and the enthalpy change is -572 kJ. Heat evolved for 1 mole of H2O \(-\) \(= \frac{-572\text{ kJ}}{2\text{ mol}}\) \(= -286\text{ kJ/mol}\)

b. Heat evolved for 4.03 g H2

Now, let's find the heat evolved when 4.03 g of hydrogen reacts with excess oxygen. First, convert grams of hydrogen to moles using the molar mass of hydrogen (2 g/mol). Moles of H2 = \(\frac{4.03 \text{g}}{2 \text{g/mol}}\) = 2.015 mol H2 Next, since the stoichiometry ratio of H2 to H2O in the balanced equation is 2:2, the moles of H2O formed will be equal to the moles of H2: Moles of H2O = 2.015 mol Now, use the heat evolved for the formation of 1 mole of H2O from part a to find the total heat evolved: Heat evolved \(-\) = \(2.015\text{ mol} \times -286\text{ kJ/mol}\) \(- = 576.29\text{ kJ}\)

c. Heat evolved for 186 g O2

Finally, let's find the heat evolved when 186 g of oxygen reacts with excess hydrogen. First, convert grams of oxygen to moles using its molar mass (32 g/mol). Moles of O2 = \(\frac{186 \text{g}}{32 \text{g/mol}}\) = 5.8125 mol O2 Since the stoichiometry ratio of O2 to H2O in the balanced equation is 1:2, the moles of H2O formed will be twice the moles of O2: Moles of H2O = 5.8125 mol × 2 = 11.625 mol H2O Now, use the heat evolved for the formation of 1 mole of H2O from part a to find the total heat evolved: Heat evolved \(-\) = \(11.625\text{ mol} \times -286\text{ kJ/mol}\) \(- = 3326.5\text{ kJ}\)

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that involves the quantitative relationship between reactants and products in a chemical reaction. In simpler terms, it's all about the 'recipe' for a chemical reaction, telling us how much of each substance is needed or produced.

For example, when we look at the reaction given in the exercise \(2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)\), stoichiometry informs us that two molecules of hydrogen gas \(\mathrm{H}_{2}\) react with one molecule of oxygen gas \(\mathrm{O}_{2}\) to produce two molecules of water \(\mathrm{H}_{2} \mathrm{O}\). This balanced equation serves as the 'recipe,' and understanding it is crucial for working out how much heat is evolved for given amounts of reactants.

Application in Heat Calculations

Applying stoichiometry to heat calculations allows us to determine the heat evolved or absorbed during a reaction. By knowing the stoichiometric coefficients (the numbers in front of the reactants and products), we can calculate the heat change for different amounts of reactants used or products formed, just as demonstrated in the exercise's solution.

Heat of Reaction

The heat of reaction, or enthalpy change (\(\Delta H\)), is the amount of heat released or absorbed when a chemical reaction takes place at constant pressure. It is an essential concept in understanding energy changes during reactions.

In the example provided, the reaction has an enthalpy change of -572 kJ, indicating that 572 kJ of heat is released when two moles of water are formed. The negative sign shows that the reaction is exothermic; it gives off heat to the surroundings. By knowing the heat of reaction, we can calculate the heat involved for various amounts of reactants or products.

Relation to Stoichiometry

Combining the concept of the heat of reaction with stoichiometry can guide us in determining how the amount of heat changes with different quantities of substances in a balanced equation. By doing this, we can solve complex real-life problems where we need to find out the energy involved in producing or consuming specific amounts of substances, such as in industrial chemical processes or understanding environmental impacts.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's a physical property that is fundamental to stoichiometry because it bridges the gap between the mass of a substance and the amount of substance (moles).

Every element has a different molar mass, based on the average mass of its atoms. For instance, in the given exercise, hydrogen has a molar mass of 2 g/mol, and oxygen has a molar mass of 32 g/mol. To find out how much heat is evolved when reacting a certain mass of reactant, such as hydrogen or oxygen, we first convert the mass to moles using their respective molar masses. This step is crucial as it allows us to properly apply the stoichiometric ratios from the balanced equation to determine the amount of product formed and consequently the heat evolved.

Importance in Calculating Heat Evolved

Without knowing the molar masses, we wouldn't be able to connect the mass of the reactants to the stoichiometric coefficients in a reaction. This connection is essential for predicting the outcome of chemical reactions and for practical applications in chemistry, such as designing reactors and calculating yields.