Chapter 5: Problem 106
Equal weights of ethane and hydrogen are mixed in an empty container at \(25^{\circ} \mathrm{C}\). The fraction to total pressure exerted by hydrogen is: (a) \(1: 2\) (b) \(1: 1\) (c) \(1: 16\) (d) \(15: 16\)
Short Answer
Expert verified
The fraction of total pressure exerted by hydrogen is \(15:16\), option (d).
Step by step solution
01
Identify Molar Masses
Identify the molar masses of ethane (C\(_2\)H\(_6\)) and hydrogen (H\(_2\)). Ethane has a molar mass of 30 g/mol, while hydrogen has a molar mass of 2 g/mol.
02
Calculate Moles of Ethane and Hydrogen
Since equal weights of ethane and hydrogen are mixed, assume a mass of \(x\) grams for each gas. The moles of ethane are \(\frac{x}{30}\) and the moles of hydrogen are \(\frac{x}{2}\).
03
Find Total Moles
Calculate the total moles in the mixture: \[ \text{Total moles} = \frac{x}{30} + \frac{x}{2} = \frac{x+15x}{30} = \frac{16x}{30} \]
04
Fraction of Total Pressure from Hydrogen
The fraction of pressure contributed by hydrogen is the ratio of its moles to the total moles, given by: \[ \frac{\frac{x}{2}}{\frac{16x}{30}} = \frac{30}{32} = \frac{15}{16} \]
05
Conclusion
Therefore, the correct fraction of total pressure contributed by hydrogen is \(\frac{15}{16}\), which corresponds to option (d).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Mass
Molar mass is an essential concept in chemistry. It relates to the mass of one mole of a substance, usually expressed in grams per mole (g/mol).
It's calculated by summing the atomic masses of the atoms that constitute a molecule.
For example, in the exercise, we looked at ethane (C\(_2\)H\(_6\)) and hydrogen (H\(_2\)). Ethane has a carbon atom (5.03 g/mol) and hydrogen (1 g/mol), leading to a molar mass calculation of \(2 \times 12 + 6 \times 1 = 30\) g/mol for ethane and \(2 \times 1= 2\) g/mol for hydrogen.
This calculation is crucial because it allows us to convert between grams and moles, a necessary step when reacting or mixing substances.
It's the bridge to understanding how much of one substance interacts with another on a molecular level.
Remember, knowing the molar mass of a compound is like having the right tools for building our understanding of chemical interactions. It's the first step in quantifying how molecules participate in chemical reactions.
It's calculated by summing the atomic masses of the atoms that constitute a molecule.
For example, in the exercise, we looked at ethane (C\(_2\)H\(_6\)) and hydrogen (H\(_2\)). Ethane has a carbon atom (5.03 g/mol) and hydrogen (1 g/mol), leading to a molar mass calculation of \(2 \times 12 + 6 \times 1 = 30\) g/mol for ethane and \(2 \times 1= 2\) g/mol for hydrogen.
This calculation is crucial because it allows us to convert between grams and moles, a necessary step when reacting or mixing substances.
It's the bridge to understanding how much of one substance interacts with another on a molecular level.
Remember, knowing the molar mass of a compound is like having the right tools for building our understanding of chemical interactions. It's the first step in quantifying how molecules participate in chemical reactions.
Moles and Stoichiometry
Moles and stoichiometry are foundational aspects of chemistry that help us understand and predict chemical reactions.
A mole is a measure of quantity, specifically \(6.022 \times 10^{23}\) of anything, typically atoms or molecules.
In the given exercise, knowing the moles of ethane and hydrogen is crucial for determining the contribution of each gas to the mixture's properties. With equal masses of each gas, the moles can be calculated using their molar masses:
- Moles of ethane are \( \frac{x}{30} \) where \(x\) is the mass.
- Moles of hydrogen are \( \frac{x}{2} \).
This stoichiometric calculation helps us determine proportions and predict the characteristics of reactions or mixtures.
Stoichiometry is the math behind chemistry, tapping into proportions and ratios that define how substances combine or decompose.
Grasping moles and stoichiometry allows you to balance equations and calculate yields. It's about understanding the big picture of tiny particles.
A mole is a measure of quantity, specifically \(6.022 \times 10^{23}\) of anything, typically atoms or molecules.
In the given exercise, knowing the moles of ethane and hydrogen is crucial for determining the contribution of each gas to the mixture's properties. With equal masses of each gas, the moles can be calculated using their molar masses:
- Moles of ethane are \( \frac{x}{30} \) where \(x\) is the mass.
- Moles of hydrogen are \( \frac{x}{2} \).
This stoichiometric calculation helps us determine proportions and predict the characteristics of reactions or mixtures.
Stoichiometry is the math behind chemistry, tapping into proportions and ratios that define how substances combine or decompose.
Grasping moles and stoichiometry allows you to balance equations and calculate yields. It's about understanding the big picture of tiny particles.
Partial Pressure
Partial pressure is a concept related to gases and is crucial when dealing with gas mixtures like in the exercise.
It's the pressure exerted by a single gas in a mixture if it occupied the whole volume of the container on its own.
By using Dalton's Law of Partial Pressures:
\[ P_{\text{total}} = P_{1} + P_{2} + 8cdots \]
we see that the total pressure is the sum of the partial pressures of individual gases.
The fraction of total pressure exerted by hydrogen, calculated as \( \frac{\frac{x}{2}}{\frac{16x}{30}} \), simplifies to \( \frac{15}{16} \) of the total pressure.
This importance lies in understanding how each gas contributes to overall pressure, especially in reactions where pressure can affect equilibrium or yields.
When dealing with gases, partial pressures inform many characteristics, including how gases behave in mixtures under various conditions and help predict movements and interactions.
It's the pressure exerted by a single gas in a mixture if it occupied the whole volume of the container on its own.
By using Dalton's Law of Partial Pressures:
\[ P_{\text{total}} = P_{1} + P_{2} + 8cdots \]
we see that the total pressure is the sum of the partial pressures of individual gases.
The fraction of total pressure exerted by hydrogen, calculated as \( \frac{\frac{x}{2}}{\frac{16x}{30}} \), simplifies to \( \frac{15}{16} \) of the total pressure.
This importance lies in understanding how each gas contributes to overall pressure, especially in reactions where pressure can affect equilibrium or yields.
When dealing with gases, partial pressures inform many characteristics, including how gases behave in mixtures under various conditions and help predict movements and interactions.
Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal.
This means that the concentrations of reactants and products remain constant over time.
In the context of gas laws, equilibrium can be influenced by factors like pressure, which is why understanding partial pressures is key.
If we change the conditions of a reaction, say by altering pressures, the position of equilibrium might shift according to Le Chatelier's Principle.
This principle posits that if an external change is applied, the system will adjust to counteract that change and re-establish equilibrium.
When we consider gas mixtures, equilibrium dynamics can drive reactions forward or reverse, depending on conditions.
Understanding equilibrium helps us control reactions, for instance, ensuring conditions are optimal for the desired products to form. Keeping an eye on how equilibrium interacts with gas laws guides our grasp of chemical processes and their potential adjustments.
This means that the concentrations of reactants and products remain constant over time.
In the context of gas laws, equilibrium can be influenced by factors like pressure, which is why understanding partial pressures is key.
If we change the conditions of a reaction, say by altering pressures, the position of equilibrium might shift according to Le Chatelier's Principle.
This principle posits that if an external change is applied, the system will adjust to counteract that change and re-establish equilibrium.
When we consider gas mixtures, equilibrium dynamics can drive reactions forward or reverse, depending on conditions.
Understanding equilibrium helps us control reactions, for instance, ensuring conditions are optimal for the desired products to form. Keeping an eye on how equilibrium interacts with gas laws guides our grasp of chemical processes and their potential adjustments.
