Imagine that the reaction $2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)$ occurs in a container that has a piston that moves to maintain a constant pressure when the reaction occurs at constant temperature. Which of the following statements describes how the volume of the container changes due to the reaction: (a) the volume increases by \(50 \%,\) (b) the volume increases by \(33 \%\), (c) the volume remains constant, (d) the volume decreases by \(33 \%,(\mathbf{e})\) the volume decreases by $50 \%\(. [Sections 10.3 and 10.4\)]$

Short Answer

Expert verified
The volume of the container decreases by approximately 33% due to the reaction, as the number of gas moles decreases from 3 to 2 moles. Therefore, the correct answer is (d).

Step by step solution

01

Calculate Moles on Both Sides of Reaction

Before the reaction, there are 2 moles of carbon monoxide (CO) and 1 mole of oxygen (O2). So, there are a total of 3 moles of gas before the reaction. After the reaction, there are 2 moles of carbon dioxide (CO2).
02

Compare Moles on Both Sides

Now, let's compare the moles of gas before and after the reaction: Moles before reaction = 3 moles Moles after reaction = 2 moles
03

Calculate Change in Volume

The number of moles of gas decreased from 3 to 2. At constant pressure and temperature, the volume of the gas is directly proportional to the number of moles. Since we observe that the moles are reduced by 1 mole, the percentage decrease in volume can be calculated by: Percentage decrease in volume = \(\frac{(initial \ moles - final \ moles)}{initial \ moles} \times 100\) Percentage decrease in volume = \(\frac{(3 - 2)}{3} \times 100\) Percentage decrease in volume = \(33.33 \% \) The volume decreases by approximately 33%. Therefore, the correct answer is (d).

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