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Chapter 15: Problem 50

For $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{2}\( at \)700 \mathrm{~K}$. In a 2.00-L vessel, the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

Short Answer

Expert verified
There are approximately 0.573 grams of SO₂ in the vessel.

Step by step solution

01

Calculate the moles of gases present

Given, 1.17 g of SO₃ and 0.105 g of O₂ are present. Use the molar mass of SO₃(80.07 g/mol) and O₂(32 g/mol) to convert grams to moles. Moles of SO₃ = (1.17 g)/(80.07 g/mol) ≈ 0.0146 mol Moles of O₂ = (0.105 g)/(32 g/mol) ≈ 0.00328 mol
02

Find the moles of SO₂ formed

Since the balanced equation states that 2 moles of SO₂ are produced for each mole of O₂ and SO₃ consumed, the change in moles can be written as: 2[s(SO₂)] = [s(O₂)] + [s(SO₃)], where s stands for the stoichiometric coefficient. Using the calculated moles, we can find the moles of SO₂ formed, 2[s(SO₂)] = 0.00328 mol + 0.0146 mol [s(SO₂)] ≈ 0.00894 mol.
03

Calculate the partial pressures of each gas

To get partial pressures, we need to use the ideal gas law which states PV = nRT. Since R and T are constants, we can express the partial pressure (P) as a function of moles (n) and volume (V). For example, P(SO₂) = [n(SO₂) × R × T] / V. Given the volume of the vessel is 2.00 L, we can calculate partial pressures: P(SO₂) = (0.00894 mol × R × 700 K) / 2.00 L P(O₂) = (0.00328 mol × R × 700 K) / 2.00 L P(SO₃) = (0.0146 mol × R × 700 K) / 2.00 L
04

Apply the equilibrium constant expression

By definition of the equilibrium constant Kₚ = \(\frac{P_{SO_3}^2}{P_{O_2} * P_{SO_2}^2}\). We know that Kₚ = 3.0 × 10² at 700K. By plugging the partial pressures from step 3 into the expression: 3.0 × 10² = \(\frac{(0.0146 * R * 700/2.00)^2}{(0.00328 * R * 700/2.00)*(0.00894 * R * 700/2.00)^2}\)
05

Solve for R

Note that R will cancel out in step 4, so the equation can be rewritten as a function of moles. 3.0 × 10² = \(\frac{(0.0146 * 700/2.00)^2}{(0.00328 * 700/2.00)*(0.00894 * 700/2.00)^2}\) Now we can solve for the moles of SO₂: Moles of SO₂ = 0.00894
06

Calculate the grams of SO₂

Finally, we can convert the moles of SO₂ back to grams using the molar mass of SO₂ (64.07 g/mol): Grams of SO₂ = moles of SO₂ × molar mass of SO₂ Grams of SO₂ ≈ 0.00894 mol × 64.07 g/mol ≈ 0.573 g. There are approximately 0.573 grams of SO₂ in the vessel.

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Most popular questions from this chapter

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and $1.50 \mathrm{~mol} \mathrm{H}_{2}\( are placed in a 3.00-L container at \)395^{\circ} \mathrm{C}$, the following reaction occurs: $\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\(. If \)K_{c}=0.802$, what are the concentrations of each substance in the equilibrium mixture?

Nitric oxide (NO) reacts readily with chlorine gas as follows: $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)$$ At \(700 \mathrm{~K},\) the equilibrium constant \(K_{p}\) for this reaction is \(2.6 \times 10^{-3}\). Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium. (a) $P_{\mathrm{NO}}=20.3 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=20.3 \mathrm{kPa}, R_{\mathrm{NOCl}}=20.3 \mathrm{kPa}$ (b) $P_{\mathrm{NO}}=25.33 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=15.2 \mathrm{kPa}, R_{\mathrm{NOCl}}=2.03 \mathrm{kPa}$ (c) $P_{\mathrm{NO}}=15.2 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=42.6 \mathrm{kPa}, P_{\mathrm{NOCl}}=5.07 \mathrm{kPa}$

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and $1.50 \mathrm{~mol} \mathrm{H}_{2}\( are placed in a 3.00-L container at \)395^{\circ} \mathrm{C}$, the following reaction occurs: $\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$. If \(K_{c}=0.802\), what are the concentrations of each substance in the equilibrium mixture?

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction $4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\( in a fetus, compared to \)K_{c}$ for the same reaction in an adult.

The following equilibria were measured at \(823 \mathrm{~K}\) : $$\begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array}$$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)$ \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K}\). (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of $101.3 \mathrm{kPa}\( and a temperature of \)298 \mathrm{~K},$ what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to $823 \mathrm{~K}$ and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?
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