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Chapter 1: Problem 19

In an experiment 2 moles of \(\mathrm{H}_{2}(g)\) and 1 mole of \(\mathrm{O}_{2}(g)\) were completely reacted, according to the following equation in a sealed container of constant volume and temperature: $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)$$ If the initial pressure in the container before the reaction is denoted as \(P_{i}\) which of the following expressions gives the final pressure, assuming ideal gas behavior? (A) \(P_{i}\) (B) 2\(P_{i}\) (C) \((3 / 2) P_{i}\) (D) \((2 / 3) P_{i}\)

Short Answer

Expert verified
The final pressure is \((2 / 3) P_{i}\), so the correct answer is (D) \((2 / 3) P_{i}\).

Step by step solution

01

Identify Initial Moles

First, identify the number of moles before the reaction begins based on the given experiment details. Here there are two moles of hydrogen gas, \(\mathrm{H}_{2}(g)\), & one mole of oxygen gas, \(\mathrm{O}_{2}(g)\). Therefore, the total initial moles, \(n_{i}\), equals 2 moles \( \mathrm{H}_{2}(g) \) + 1 mole \( \mathrm{O}_{2}(g) \) = 3 moles.
02

Identify Final Moles

Next, identify the number of moles after the reaction. According to the balanced chemical equation, two moles of \(\mathrm{H}_{2}\mathrm{O}(g)\) are formed by completely reacting two moles of \(\mathrm{H}_{2}(g)\) and one mole of \(\mathrm{O}_{2}(g)\). Thus, the total final moles, \(n_{f}\), equals 2 moles \(\mathrm{H}_{2}\mathrm{O}(g)\).
03

Calculate Ratio of Final to Initial Moles

Now, calculate the ratio of final to initial moles, \(\frac{n_{f}}{n_{i}}\) , which equals 2 / 3 = \frac{2}{3} .
04

Apply Ideal Gas Law

Since the volume and temperature of the container are constant before and after the reaction, the ratio \(\frac{n_{f}}{n_{i}}\) is equal to the ratio of the final pressure to the initial pressure, \(\frac{P_{f}}{P_{i}}\), because of the direct proportionality between pressure and the number of moles, according to the ideal gas law. Therefore, \(\frac{P_{f}}{P_{i}}\) equals \(\frac{2}{3}\). Solve the equation \(P_{f} = \(\frac{2}{3}\) P_{i}\) for \(P_{f}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
A chemical reaction involves the rearrangement of atoms to transform reactants into products. In the given problem, we deal with a specific reaction: the combustion of hydrogen gas
  • The reaction equation is: \(2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(g)\).
  • This equation tells us that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water vapor.
  • A balanced chemical equation is crucial because it provides the ratio of moles involved in the reaction, which helps us understand the stoichiometry.
Understanding these ratios allows us to predict how much of each substance is consumed and produced in the reaction. This example shows how a simple reaction can lead to significant pressure changes in a container if the gas is ideal.
Mole Concept
The mole concept is a fundamental aspect of understanding chemical reactions. It helps quantify the amount of substance involved in a reaction. A mole represents Avogadro's number \(6.022 \times 10^{23}\) of particles, be it atoms, molecules, or ions. In this specific problem, it helps in determining the number of molecules participating in the reaction:
  • Initially, there are three moles of reactants: two moles of \(\mathrm{H}_{2}(g)\) and one mole of \(\mathrm{O}_{2}(g)\).
  • After the reaction, two moles of \(\mathrm{H}_{2}\mathrm{O}(g)\) are produced.
The concept highlights the importance of mole ratios derived from the balanced chemical equations, aiding in the calculation of other properties like gas volume and pressure.
Pressure Changes
Pressure changes in a gas-filled container depend significantly on the number of gaseous moles before and after the reaction takes place. In this closed system with constant volume and temperature:
  • The initial pressure, \(P_{i}\), is affected by the initial moles of gas.
  • After the reaction, fewer moles of gas remain due to the creation of water vapor, which changes the pressure.
  • The pressure drops because the number of gaseous moles decreases from 3 moles to 2 moles.
This decrease in pressure highlights the direct proportionality between the number of moles and pressure in a gas, as explained by the ideal gas law. Pressure changes are essential for predicting system behaviors in chemical reactions.
Stoichiometry
Stoichiometry leverages the coefficients of a balanced chemical equation to relate quantities of reactants and products. In our equation where \(2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(g)\), we can observe important stoichiometric relationships:
  • The stoichiometric coefficients indicate that two moles of \(\mathrm{H}_{2}(g)\) and one mole of \(\mathrm{O}_{2}(g)\) produce two moles of \(\mathrm{H}_{2}\mathrm{O}(g)\).
  • This equation provides the ratio \( \frac{2 \, \text{moles of } \mathrm{H}_{2}\mathrm{O}(g)}{3 \, \text{moles of reactants}} = \frac{2}{3} \).
  • This ratio directly influences the calculation of final pressure in the gas container.
Such calculations are fundamental in predicting the outcomes of chemical reactions, including changes in pressure, as demonstrated in the problem.

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Most popular questions from this chapter

A student added 1 liter of a 1.0\(M\) \(\mathrm{KCl}\) solution to 1 liter of a 1.0 \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution. A lead chloride precipitate formed, and nearly all of the lead ions disappeared from the solution. Which of the following lists the ions remaining in the solution in order of decreasing concentration? (A) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]\) (B) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{K}^{+}\right]\) (C) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{NO}_{3}^{-}\right]\) (D) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]\)

A bottle of water is left outside early in the morning. The bottle warms gradually over the course of the day. What will happen to the pH of the water as the bottle warms? (A) Nothing; pure water always has a pH of 7.00. (B) Nothing; the volume would have to change in order for any ion concentration to change. (C) It will increase because the concentration of \(\left[\mathrm{H}^{+}\right]\) is increasing. (D) It will decrease because the auto-ionization of water is an endothermic process.

Which substance would have the highest boiling point? (A) Ethanol, because it is the most asymmetrical (B) Acetone, because of the double bond (C) Ethylene glycol, because it has the most hydrogen bonding (D) All three substances would have very similar boiling points because their molar masses are similar.

\(2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \leftrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) \Delta H=167 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) During the reaction above, the product yield can be increased by increasing the temperature of the reaction. Why is this effective? (A) The reaction is endothermic; therefore adding heat will shift it to the right. (B) Increasing the temperature increases the speed of the molecules, meaning there will be more collisions between them. (C) The reactants are less massive than the products, and an increase in temperature will cause their kinetic energy to increase more than that of the products. (D) The increase in temperature allows for a higher percentage of molecular collisions to occur with the proper orientation to create the product.

$$2 \mathrm{NOCl} \rightarrow 2 \mathrm{NO}+\mathrm{Cl}_{2}$$ The reaction above takes place with all of the reactants and products in the gaseous phase. Which of the following is true of the relative rates of disappearance of the reactants and appearance of the products? (A) NO appears at twice the rate that NOCl disappears. (B) NO appears at the same rate that NOCl disappears. (C) NO appears at half the rate that NOCl disappears. (D) \(\mathrm{Cl}_{2}\) appears at the same rate that NOCl disappears.
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