Question

When acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), is burned in oxygen, carbon dioxide and steam are formed. A sample of acetylene with a volume of \(7.50 \mathrm{~L}\) and a pressure of \(1.00\) atm is burned in excess oxygen at \(225^{\circ} \mathrm{C}\). The products are transferred without loss to a 10.0-L flask at the same temperature. (a) Write a balanced equation for the reaction. (b) What is the total pressure of the products in the \(10.0\) - \(\mathrm{L}\) flask? (c) What is the partial pressure of each of the products in the flask?

Short answer

Answer: (a) The balanced equation for the reaction is \(\mathrm{C}_{2} \mathrm{H}_{2} + \frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\). (b) The total pressure of the products in the 10.0-L flask is 2.23 atm. (c) The partial pressure of CO2 is 1.48 atm, and the partial pressure of H2O is 0.74 atm.

Step-by-step solution

Balanced equation for the reaction

The balanced equation for the reaction of acetylene with oxygen is given by: \(\mathrm{C}_{2} \mathrm{H}_{2} + \frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\) Next, we'll use the Ideal Gas Law to determine the number of moles of acetylene in the given conditions (7.50 L, 1.00 atm, and \(225^{\circ} \mathrm{C}\)).

Ideal Gas Law

The Ideal Gas Law equation is given by \(PV = nRT\), where: - P is the pressure (1.00 atm) - V is the volume (7.50 L) - n is the number of moles - R is the gas constant (0.0821 L⋅atm/mol⋅K) - T is the temperature in Kelvin (225+273.15=498.15 K) Rearrange the equation to solve for n: \(n = \frac{PV}{RT}\) Plug in the given values and solve for n: \(n = \frac{(1.00 \ \mathrm{atm})(7.50 \ \mathrm{L})}{(0.0821 \ \mathrm{L \cdot atm/mol \cdot K})(498.15 \ \mathrm{K})} = 0.182 \ \mathrm{moles}\) Now, we'll use stoichiometry to find the moles of CO2 and H2O produced from the reaction.

Stoichiometry

From the balanced equation, the stoichiometric coefficients are: - Acetylene: 1 - Oxygen: 2.5 - Carbon dioxide: 2 - Water: 1 From the number of acetylene moles we just calculated, we can now determine the moles of carbon dioxide and water: - Moles of CO2 produced: \(0.182 \cdot 2 = 0.364 \ \mathrm{moles}\) - Moles of H2O produced: \(0.182 \cdot 1 = 0.182 \ \mathrm{moles}\) Now, we'll determine the total pressure and the partial pressures of the products in the \(10.0\)-L flask.

Total Pressure

The Ideal Gas Law equation is still given by \(PV = nRT\). We can use this equation to determine the total pressure of the products in the \(10.0\)-L flask. To do this, we'll find the total number of moles of the products (CO2 and H2O), the given volume (10.0 L), and the temperature (498.15 K). - Total moles of products: \(0.364 + 0.182 = 0.546 \ \mathrm{moles}\) - Volume: \(10.0 \ \mathrm{L}\) - Temperature: \(498.15 \ \mathrm{K}\) Now substitute these values back into the Ideal Gas Law equation and solve for the total pressure of the products: \(P = \frac{nRT}{V} = \frac{(0.546 \ \mathrm{moles})(0.0821 \ \mathrm{L \cdot atm/mol \cdot K})(498.15 \ \mathrm{K})}{10.0 \ \mathrm{L}}= 2.23 \ \mathrm{atm}\) Finally, let's determine the partial pressures of CO2 and H2O in the flask.

Partial Pressures

Using Dalton's Law of Partial Pressures, the partial pressures of CO2 and H2O can be calculated as: - Partial pressure of CO2: \(P_{CO2} = \frac{\mathrm{moles \ of \ CO2}}{\mathrm{total \ moles \ of \ products}} \times \mathrm{total \ pressure} = \frac{0.364}{0.546} \times 2.23 \ \mathrm{atm} = 1.48 \ \mathrm{atm}\) - Partial pressure of H2O: \(P_{H2O} = \frac{\mathrm{moles \ of \ H2O}}{\mathrm{total \ moles \ of \ products}} \times \mathrm{total \ pressure} = \frac{0.182}{0.546} \times 2.23 \ \mathrm{atm} = 0.74 \ \mathrm{atm}\) In conclusion, (a) The balanced equation for the reaction is \(\mathrm{C}_{2} \mathrm{H}_{2} + \frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\). (b) The total pressure of the products in the \(10.0\) - L flask is \(2.23\) atm. (c) The partial pressure of CO2 is \(1.48\) atm, and the partial pressure of H2O is \(0.74\) atm.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry dealing with the quantitative relationships between reactants and products in a chemical reaction. It's like baking a cake: you need specific amounts of ingredients to get a perfect cake. In chemical reactions, stoichiometry uses the coefficients from balanced chemical equations to calculate the amount of product formed from a given amount of reactant.

In the reaction of acetylene with oxygen, understanding stoichiometry is crucial:

• The balanced chemical equation, \[ \mathrm{C}_{2} \mathrm{H}_{2} + \frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O} \] shows us that one mole of acetylene reacts with 2.5 moles of oxygen to produce two moles of carbon dioxide and one mole of water.
• By knowing the moles of acetylene (0.182 moles) in the reaction, stoichiometry allows us to calculate the moles of carbon dioxide and water formed.
• For every one mole of acetylene, two moles of CO2 and one mole of water are produced, leading to the formation of 0.364 moles of CO2 and 0.182 moles of water from 0.182 moles of acetylene.

Stoichiometry helps us understand and predict the yields of reactions, ensuring we know exactly how much of each substance we'll have after the reaction is complete.

Balanced Chemical Equation

A balanced chemical equation is a representation of a chemical reaction where the number of atoms for each element is equal on both the reactant and product sides. It's essential because it reflects the conservation of mass, meaning atoms are neither created nor destroyed in a chemical reaction.

Let's consider the equation:

• \[ \mathrm{C}_{2} \mathrm{H}_{2} + \frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O} \]

Here’s why balancing equations is important:

• Conservation of Mass: The total number of atoms of each element involved must be the same before and after the reaction, demonstrating that mass is conserved.
• Stoichiometric Relations: The coefficients (numbers in front of molecules) indicate the precise ratio of reactants that will react together and the ratio of products formed. For instance, for complete combustion of acetylene, you need 2.5 moles of oxygen for every mole of acetylene.
• Predictive Power: Balancing gives us the exact proportions allowing for predictions about the volumes or masses of gases produced or consumed in reactions under given conditions, crucial for industrial chemical processes.

Learning to balance equations assures you are correctly understanding and depicting chemical reactions.

Partial Pressure

Partial pressure is a concept from Dalton’s Law which states that in a mixture of gases, each gas exerts a pressure independently of the others as if it were present alone in the container. It's particularly useful here since the reaction produces a mixture of gases.

In our reaction, calculating partial pressure involves:

• Recognizing each gas's contribution to the total pressure within a gas mixture. The total pressure is the sum of the partial pressures of all gases present.
• Using the formula for partial pressure, \[ P_{\text{partial}} = \frac{\text{moles of gas}}{\text{total moles}} \times \text{total pressure} \]
• Determining the partial pressures from the reaction products: Given \(2.23 \text{ atm}\) total pressure and knowing moles of each gas (CO2 and H2O), their partial pressures are calculated as \(1.48 \text{ atm}\) for CO2 and \(0.74 \text{ atm}\) for H2O.

Understanding partial pressures helps in determining how each gas within a mixture contributes to the overall conditions, such as pressure and volume in closed systems. This concept is essential in fields ranging from atmospheric science to designing chemical reactors.