Problem 99

Question

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(8.60 \mathrm{g} \mathrm{O}_{2}\) is introduced into a 1.500 L flask at \(25^{\circ} \mathrm{C}\). When the mixture is ignited, an explosive reaction occurs in which water is the only product. What is the total gas pressure when the flask is returned to \(25^{\circ} \mathrm{C} ?\) (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8 \mathrm{mmHg}\).)

Step-by-Step Solution

Verified
Answer
The total gas pressure in the flask at the end of the reaction is approximately \(1.888 atm\).
1Step 1: Balanced Chemical equation
From the statement, Hydrogen (\(H_{2}\)) reacts with Oxygen (\(O_{2}\)) to form water (\(H_{2}O\)). The balanced equation becomes: \(2H_{2} + O_{2} \rightarrow 2H_{2}O\).
2Step 2: Find Moles of Reactants
The amount of substance can be calculated using the formula n = m / Mr. Hence, the number of moles for: Hydrogen (\(H_{2}\)) is \(1.00 g / 2.02 g/mol \approx 0.495 mol\) and Oxygen (\(O_{2}\)) is \(8.60 g / 32.00 g/mol \approx 0.269 mol\).
3Step 3: Find Moles of Products
From the stoichiometry of the chemical equation, it shows that the amount of \(H_{2}O\) formed is twice the amount of \(H_{2}\) used. Therefore, \(0.495 mol / 2 = 0.2475 mol\) of \(H_{2}O\) vapor is produced. Consequently, \(O_{2} = 0.269 mol - 0.2475 mol = 0.0215 mol\) is unused and remains.
4Step 4: Use Ideal Gas Law to find Pressure
The Ideal Gas Law is given by \(PV = nRT\), where P is pressure, V is volume, n is the number of moles, R is the gas constant and T is the temperature in Kelvin. For this calculation, R = 0.0821 Latm/molK and T = 25°C + 273.15 = 298.15 K. Solving for pressure, \(P = nRT/V = (0.0215 mol + 0.2475 mol)(0.0821 Latm/molK)(298.15K) / 1.500 L = 1.919 atm\).
5Step 5: Subtract Water Vapor Pressure
The total pressure earlier calculated includes the pressure due to water vapor. The water vapor pressure must be subtracted from the total pressure to get the final pressure in the flask. The pressure of water vapor at 25°C is given as 23.8 mmHg, which can be converted to atm by using the conversion factor 1 atm = 760 mmHg. Hence, 23.8 mmHg = \(23.8 mmHg / 760 = 0.0313 atm\). The final pressure in the flask will be \(1.919 atm - 0.0313 atm = 1.888 atm\) rounded to three decimal places.

Key Concepts

StoichiometryChemical ReactionVapor PressureMole Calculation
Stoichiometry
Stoichiometry is the part of chemistry that deals with the quantitative relationships between the substances involved in chemical reactions. It allows us to predict how much of each substance is consumed or produced in a reaction.
In the balanced equation \(2H_{2} + O_{2} \rightarrow 2H_{2}O\), stoichiometry tells us that every 2 moles of hydrogen \(H_{2}\) react with 1 mole of oxygen \(O_{2}\) to form 2 moles of water \(H_{2}O\). By using stoichiometry, we can determine how much oxygen is required or how much water will form from a given amount of hydrogen.
This is vital in understanding the calculations and ensuring that the reaction conditions are optimal.
Chemical Reaction
A chemical reaction involves the rearrangement of atoms to form new substances. In this problem, hydrogen gas reacts with oxygen gas to form water.
The reaction is explosive because it's highly exothermic, meaning it releases a lot of energy. Reactants \(H_{2}\) and \(O_{2}\) rearrange during ignition to form \(H_{2}O\) molecules.
The equation \(2H_{2} + O_{2} \rightarrow 2H_{2}O\) represents this process, where the reactants on the left side convert to products on the right. Understanding these basics helps in visualizing how substances transform during chemical reactions.
Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid form at a given temperature. It is a crucial concept in calculating the final pressure of a gas mixture.
In this exercise, after the reaction forms water as a product, its vapor exerts pressure within the flask. The given vapor pressure of water at 25°C is 23.8 mmHg.
This needs to be subtracted from the total pressure to understand the pressure solely due to the remaining gases. Converting this to atm using the conversion 1 atm = 760 mmHg is essential for accurate calculations and involves knowing how vapor affects gas mixtures.
Mole Calculation
Mole calculation is a central concept in chemistry that helps quantify substances. It involves converting mass to moles using molar mass, and vice versa.
In this problem, you calculate moles of hydrogen \(H_{2}\) as \(1.00 \text{ g} / 2.02 \text{ g/mol} \approx 0.495 \text{ mol}\) and oxygen \(O_{2}\) as \(8.60 \text{ g} / 32.00 \text{ g/mol} \approx 0.269 \text{ mol}\).
Accurate mole calculation enables precise stoichiometric predictions, such as determining excess reactants or calculating product formation. It's integral to connecting masses, moles, and reactions, linking the physical and chemical worlds.