Question

(a) Consider the combustion of hydrogen, \(2 \mathrm{H}_{2}(g)+\) \(\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) .\) If hydrogen is burning at the rate of 0.48 \(\mathrm{mol} / \mathrm{s}\) , what is the rate of consumption of oxygen? What is the rate of formation of water vapor? (b) The reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) is carried out in a closed vessel. If the partial pressure of \(\mathrm{NO}\) is decreasing at the rate of 56 torr/min, what is the rate of change of the total pressure of the vessel?

Short answer

(a) The rate of oxygen consumption is \(0.24 \ mol/s\), and the rate of water vapor formation is \(0.48 \ mol/s\). (b) The rate of change of the total pressure of the vessel is \(-28 \ torr/min\).

Step-by-step solution

Understand the given values

The given balanced chemical equation for the combustion of hydrogen is: \(2 H_{2}(g) + O_{2}(g) \longrightarrow 2 H_{2}O(g)\) It is given that hydrogen is burning at a rate of 0.48 mol/s.

Calculate the rate of O2 consumption

Using the stoichiometry from the balanced equation, we see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O. So, the rate at which O2 is consumed is half the rate at which H2 is consumed. Therefore, the rate of O2 consumption is: \(\frac{0.48 \ mol/s}{2} = 0.24 \ mol/s \)

Calculate the rate of H2O formation

Again, using stoichiometry, we see that for every 2 moles of H2 consumed, 2 moles of H2O are produced. Since the rate of H2 consumption is 0.48 mol/s, the rate of H2O formation will be the same: \(0.48 \ mol/s\) (a) Solution Summary: The rate of oxygen consumption is 0.24 mol/s, and the rate of water vapor formation is 0.48 mol/s. (b)

Understand the given values

The given balanced chemical equation for the reaction is: \(2 NO(g) + Cl_{2}(g) \longrightarrow 2 NOCl(g)\) It is given that the partial pressure of NO is decreasing at the rate of 56 torr/min.

Calculate the rate of Cl2 consumption

From the balanced equation, we see that 2 moles of NO react with 1 mole of Cl2 to produce 2 moles of NOCl. So, the rate at which Cl2 is consumed is half the rate at which NO is consumed. Therefore, the rate of Cl2 consumption is: \(\frac{56 \ torr/min}{2} = 28 \ torr/min\)

Calculate the rate of NOCl formation

Following the stoichiometry of the balanced equation, we see that for every 2 moles of NO consumed, 2 moles of NOCl are produced. Since the rate of NO consumption is 56 torr/min, the rate of NOCl formation will be the same: \(56 \ torr/min\)

Calculate the rate of change of the total pressure

The net rate of change in total pressure is based on the formation of one product (NOCl) as well as the consumption of two reactants (NO and Cl2). The rate of change in total pressure is given by: \(\textrm{Rate of change of total pressure} = \textrm{Rate of NOCl formation} - (\textrm{Rate of NO consumption} + \textrm{Rate of Cl2 consumption}) \) Plugging in the values, we get: \(\textrm{Rate of change of total pressure} = 56 \ torr/min - (56 \ torr/min + 28 \ torr/min) = -28 \ torr/min \) (b) Solution Summary: The rate of change of the total pressure of the vessel is -28 torr/min.

Key concepts

Stoichiometry in Chemical Reactions

Stoichiometry is the mathematical relationship between reactants and products in a chemical reaction. It is a cornerstone in the study of chemistry, as it allows us to predict the amounts of substances consumed and produced during reactions.

In the combustion of hydrogen example (\(2 \textrm{H}_2(g) + \textrm{O}_2(g) \rightarrow 2 \textrm{H}_2\textrm{O}(g)\)), the stoichiometric coefficients indicate that 2 moles of hydrogen gas will react with 1 mole of oxygen gas to produce 2 moles of water vapor. Understanding these ratios is crucial when calculating the rate at which each reactant is used and each product is formed.

For instance, if hydrogen is burning at a rate of 0.48 mol/s, we can use stoichiometry to deduce that the rate of oxygen consumption will be half of that, due to the 2:1 reaction ratio, while the rate of water vapor formation will be identical to the hydrogen consumption rate, fulfilling the 1:1 formation ratio in the reaction. By mastering stoichiometry, students can solve a variety of problems related to reaction compositions and predict outcomes based on reaction equations.

The Significance of Reaction Rates

The rate of a chemical reaction describes how fast reactants are transformed into products. It has immense practical importance in industries where controlling the speed of a reaction is crucial for efficiency and safety.

In the examples provided, the reaction rates are given for the consumption of reactants and formation of products. The rate of the reaction can be influenced by factors such as temperature, concentration of reactants, and the presence of a catalyst. For example, in the given exercise, when we determine the rate of oxygen consumption and water production, we are directly applying the concept of chemical kinetics to relate to reactants and products.

Relating Reaction Rates to Stoichiometry

By relating reaction rates to the stoichiometry of a balanced equation, as seen in the exercises, students can quantify how changes in one substance's rate affect another. It's important to always consider the coefficients of the balanced equation to make these calculations. For example, a coefficient of 2 for hydrogen in the given reaction means its rate of reaction directly influences the rate of other substances involved, like oxygen and water vapor.

Gas Laws and Pressure Changes

Gas laws describe how pressure, volume, and temperature of a gas relate to each other and can be used to predict how these properties change in response to various conditions.

In the reaction involving nitrogen monoxide and chlorine gas (\(2 \textrm{NO}(g) + \textrm{Cl}_2(g) \rightarrow 2 \textrm{NOCl}(g)\)), the change in partial pressures and the total pressure in the vessel can be understood through principles like Dalton's Law of Partial Pressures. Dalton's Law states that the total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component.

Also, the Ideal Gas Law (\(PV = nRT\)), where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature, provides a foundation for understanding how gases behave under different conditions. In the second part of the exercise, the decrease in partial pressure of \textrm{NO} and the corresponding decrease in total pressure can be analyzed through these gas law principles, connecting the behavior of gaseous substances directly to the stoichiometric relationships established by the balanced chemical equation.