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Chapter 1: Problem 18

A gaseous mixture at \(25^{\circ} \mathrm{C}\) contained 1 mole of \(\mathrm{CH}_{4}\) and 2 moles of \(\mathrm{O}_{2}\) and the pressure was measured at 2 \(\mathrm{atm}\) . The gases then underwent the reaction shown below. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) What was the pressure in the container after the reaction had gone to completion and the temperature was allowed to return to \(25^{\circ} \mathrm{C} ?\) (A) 1 atm (B) 2 \(\mathrm{atm}\) (C) 3 \(\mathrm{atm}\) (D) 4 \(\mathrm{atm}\)

Short Answer

Expert verified
The pressure in the container after the reaction completed and the temperature returned to \(25^{\circ} \mathrm{C}\) is 2 \(\mathrm{atm}\) (option B).

Step by step solution

01

Identify reaction stoichiometry

Examine the balanced chemical reaction. Every 1 mole of \(\mathrm{CH}_{4}\) reacts with 2 moles of \(\mathrm{O}_{2}\) to produce 1 mole of \(\mathrm{CO}_{2}\) and 2 moles of \(\mathrm{H}_{2}O\). The given conditions show that there's enough \(\mathrm{O}_{2}\) to react with all the \(\mathrm{CH}_{4}\), so all reactants will be consumed in the reaction.
02

Calculate the total moles after reaction

After the reaction, from the stoichiometry, 3 moles of products are produced (1 mole of \(\mathrm{CO}_{2}\) and 2 moles of \(\mathrm{H}_{2}O\). So, total moles \(n_{total}\) in the gaseous mixture after the reaction are 3.
03

Apply the Ideal Gas Law

Since temperature is constant and volume of the container doesn't change, pressure is directly proportional to the number moles, using the equation \(P_1\/n_1 = P_2\/n_2\), where \(P\) is pressure and \(n\) is number of moles. Substituting the given values, 2 \(\mathrm{atm}\) for \(P_1\) and 3 for \(n_1\) (moles before reaction), 3 for \(n_2\) (moles after reaction). Therefore, the pressure after the reaction \(P_2\) = \(P_1\) \(\times (\frac{n_2}{n_1}) = 2 \times (\frac{3}{3}) = 2 \(\mathrm{atm}\).

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Most popular questions from this chapter

$2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \leftrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) \Delta H=167 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}$ 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.

Nitrogen’s electronegativity value is between those of phosphorus and oxygen. Which of the following correctly describes the relationship between the three values? (A) The value for nitrogen is less than that of phosphorus because nitrogen is larger, but greater than that of oxygen because nitrogen has a greater effective nuclear charge. (B) The value for nitrogen is less than that of phosphorus because nitrogen has fewer protons, but greater than that of oxygen because nitrogen has fewer valence electrons. (C) The value for nitrogen is greater than that of phosphorus because nitrogen has fewer electrons, but less than that of oxygen because nitrogen is smaller. (D) The value for nitrogen is greater than that of phosphorus because nitrogen is smaller, but less than that of oxygen because nitrogen has a smaller effective nuclear charge.

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ Correctly identify the anode and cathode in this reaction as well as where oxidation and reduction are taking place. (A) Cu is the anode where oxidation occurs, and Zn is the cathode where reduction occurs. (B) Cu is the anode where reduction occurs, and Zn is the cathode where oxidation occurs. (C) Zn is the anode where oxidation occurs, and Cu is the cathode where reduction occurs. (D) Zn is the anode where reduction occurs, and Cu is the cathode where oxidation occurs.

Starting with a stock solution of 18.0 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) , what is the proper procedure to create a 1.00 \(\mathrm{L}\) sample of a 3.0 \(\mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in a volumetric flask? (A) Add 167 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (B) Add 600 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (C) Fill the flask partway with water, then add 167 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water. (D) Fill the flask partway with water, then add 600 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water.

\(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightarrow 2 \mathrm{HF}(g)\) Gaseous hydrogen and fluorine combine in the reaction above to form hydrogen fluride with an enthalpy change of \(-540 \mathrm{kJ}\) . What is the value of the heat of formation of \(\mathrm{HF}(g) ?\) (A) \(-1,080 \mathrm{kJ} / \mathrm{mol}\) (B) \(-270 \mathrm{kJ} / \mathrm{mol}\) (C) 270 \(\mathrm{kJ} / \mathrm{mol}\) (D) 540 \(\mathrm{kJ} / \mathrm{mol}\)
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