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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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\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$
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