Problem 25
Question
(a) Consider the combustion of \(\mathrm{H}_{2}(g): 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) \(\longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) .\) 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 \(/ \mathrm{min}\), what is the rate of change of the total pressure of the vessel?
Step-by-Step Solution
Verified Answer
The rate of consumption of oxygen is \(0.24 \, mol/s\) and the rate of formation of water vapor is \(0.48 \, mol/s\). The rate of change of total pressure of the vessel is \(-28 \, torr/min\).
1Step 1: Identify the stoichiometric coefficients
In the balanced combustion of hydrogen, we have the equation as:
\(2 H_{2}(g) + O_{2}(g) \longrightarrow 2 H_{2}O(g)\)
Here, the stoichiometric coefficients are 2 for hydrogen, 1 for oxygen, and 2 for water vapor.
2Step 2: Relate the rates using stoichiometry
We can relate the rate of consumption of hydrogen, oxygen, and formation of water vapor using their stoichiometric coefficients:
\(\frac{d[H_2]}{dt} : \frac{d[O_2]}{dt} : \frac{d[H_2O]}{dt} = -2 : -1 : 2\)
3Step 3: Calculate the rate of consumption of oxygen
The given rate of consumption of hydrogen is 0.48 mol/s:
\(\frac{d[H_2]}{dt} = -0.48 \, mol/s\)
Using the ratio from Step 2, we can calculate the rate of consumption of oxygen:
\(\frac{d[O_2]}{dt} = (-0.48 * (-1))/(-2) = 0.24 \, mol/s\)
4Step 4: Calculate the rate of formation of water vapor
Similarly, using the ratio from Step 2, we can calculate the rate of formation of water vapor:
\(\frac{d[H_2O]}{dt} = (-0.48 * 2)/(-2) = 0.48 \, mol/s\)
b) Rate of change of the total pressure of the vessel
5Step 1: Identify the stoichiometric coefficients
In the balanced reaction, we have the equation as:
\(2 NO(g) + Cl_2(g) \longrightarrow 2 NOCl(g)\)
Here, the stoichiometric coefficients are 2 for NO, 1 for Cl₂, and 2 for NOCl.
6Step 2: Relate the rates using stoichiometry
We can relate the rate of consumption of NO, Cl₂, and formation of NOCl using their stoichiometric coefficients:
\(\frac{d[NO]}{dt} : \frac{d[Cl_2]}{dt} : \frac{d[NOCl]}{dt} = -2 : -1 : 2\)
7Step 3: Calculate the rate of consumption of Cl₂ and formation of NOCl
The given rate of decrease in partial pressure of NO is 56 torr/min:
\(\frac{d[P_{NO}]}{dt} = -56 \, torr/min\)
Using the ratio from Step 2, we can calculate the rate of consumption of Cl₂ and formation of NOCl:
\(\frac{d[P_{Cl_2}]}{dt} = (-56 * (-1))/(-2) = -28 \, torr/min\)
\(\frac{d[P_{NOCl}]}{dt} = (-56 * 2)/(-2) = 56 \, torr/min\)
8Step 4: Calculate the rate of change of the total pressure
Now, we can calculate the rate of change of the total pressure of the vessel by adding the rates of change in partial pressures:
\(\frac{d[P_{total}]}{dt} = \frac{d[P_{NO}]}{dt} + \frac{d[P_{Cl_2}]}{dt} + \frac{d[P_{NOCl}]}{dt} = -56 \, torr/min - 28 \, torr/min + 56 \, torr/min = -28 \, torr/min\)
So, the rate of change of the total pressure of the vessel is -28 torr/min.
Key Concepts
Combustion ReactionsStoichiometryReaction Rate Calculations
Combustion Reactions
Combustion reactions are chemical processes where a substance reacts quickly with oxygen, releasing energy in the form of light or heat. They are commonly known as burning and are fundamental in everyday life, from engines to candles. In these reactions, oxygen (O₂) combines with a fuel (like hydrogen in the given exercise), to form combustion products.
For the exercise, we look at hydrogen combustion:
These equations also allow for the calculation of unknown rates. For example, if H₂ is consumed at 0.48 mol/s, you can determine the rate at which O₂ is consumed (0.24 mol/s) and the rate of water vapor formation (0.48 mol/s). This is essential for engineering and scientific calculations related to energy release and resource consumption.
For the exercise, we look at hydrogen combustion:
- Reactants: 2H₂(g) + O₂(g)
- Products: 2H₂O(g)
These equations also allow for the calculation of unknown rates. For example, if H₂ is consumed at 0.48 mol/s, you can determine the rate at which O₂ is consumed (0.24 mol/s) and the rate of water vapor formation (0.48 mol/s). This is essential for engineering and scientific calculations related to energy release and resource consumption.
Stoichiometry
Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. In simple terms, it’s all about measuring the correct amount of ingredients in chemistry, much like a recipe in cooking. Stoichiometry is crucial for understanding and balancing chemical equations.
In the context of the provided exercise, stoichiometry helps us determine the rates of consumption of reactants and formation of products. Let's go through the steps using the given reaction:
Stoichiometry not only aids in calculating chemical quantities but also ensures efficient use of resources and predictability in chemical processes.
In the context of the provided exercise, stoichiometry helps us determine the rates of consumption of reactants and formation of products. Let's go through the steps using the given reaction:
- The balanced equation: 2H₂(g) + O₂(g) → 2H₂O(g)
- The stoichiometry of the equation is 2:1:2, meaning every 2 moles of hydrogen will react with 1 mole of oxygen to produce 2 moles of water vapor.
Stoichiometry not only aids in calculating chemical quantities but also ensures efficient use of resources and predictability in chemical processes.
Reaction Rate Calculations
Reaction rate calculations are used to determine the speed of a reaction, that is, how quickly reactants are converted into products. In chemical kinetics, the reaction rate can provide insights into the dynamics of a chemical process, crucial for industries and research.
For example, let's consider the exercise with nitric oxide (NO) and chlorine (Cl₂):
Calculate it by summing the rates: i.e., i.e., -56 (NO) + -28 (Cl₂) + 56 (NOCl) which equals -28 torr/min.
Reaction rate calculations allow prediction and control of reaction speeds, essential for both safety and efficiency in chemical manufacturing.
For example, let's consider the exercise with nitric oxide (NO) and chlorine (Cl₂):
- The reaction: 2NO(g) + Cl₂(g) → 2NOCl(g)
- Here, if the partial pressure of NO decreases by 56 torr/min, we use stoichiometry to find the rate of pressure change for Cl₂ and NOCl.
- The rate of pressure decrease for Cl₂ is 28 torr/min.
- The rate of pressure increase for NOCl is 56 torr/min.
Calculate it by summing the rates: i.e., i.e., -56 (NO) + -28 (Cl₂) + 56 (NOCl) which equals -28 torr/min.
Reaction rate calculations allow prediction and control of reaction speeds, essential for both safety and efficiency in chemical manufacturing.
Other exercises in this chapter
Problem 23
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