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Chapter 12: Problem 73

Consider the decomposition of the compound \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) as follows:$$\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g)$$.When a 5.63 -g sample of pure \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g)\) was sealed into an otherwise empty 2.50 -L flask and heated to \(200 .^{\circ} \mathrm{C},\) the pressure in the flask gradually rose to 1.63 atm and remained at that value. Calculate \(K\) for this reaction.

Short Answer

Expert verified
The equilibrium constant (K) for the decomposition reaction of C5H6O3 at 200°C is approximately 9.48.

Step by step solution

01

Write the balanced chemical equation and the expression for the equilibrium constant

The balanced chemical equation for the decomposition of C5H6O3 is given as: \[C_5H_6O_3 (g) \rightarrow C_2H_6 (g) + 3 CO (g)\] The expression for the equilibrium constant K can be written as: \[K = \frac{[C_2H_6][CO]^3}{[C_5H_6O_3]}\]
02

Calculate the initial moles of C5H6O3

We are given that a 5.63-g sample of pure C5H6O_3 was sealed into the flask. To calculate the initial moles of C5H6O3, we use its molar mass, which is: \[\text{Molar mass of } C_5H_6O_3 = 5(12.01) + 6(1.01) + 3(16.00) = 60.05 + 6.06 + 48.00 = 114.11\, g/mol\] Moles of C5H6O3 = \(\frac{\text{mass of C5H6O3}}{\text{molar mass of C5H6O3}}\) Moles of C5H6O3 = \(\frac{5.63\, g}{114.11\, g/mol} = 0.0493\,mol\)
03

Calculate the initial concentrations of each species

Next, we need to calculate the initial concentrations. We can do this by dividing the moles by the volume of the flask (2.50 L). Initial concentration of C5H6O3 = \(\frac{0.0493\,mol}{2.50\,L} = 0.0197\,M\) Since no products are present initially, their initial concentrations are 0: Initial concentration of C2H6 = Initial concentration of CO = 0
04

Use the stoichiometry of the reaction to calculate the equilibrium concentrations

Since the moles of C2H6 and 3CO produced are in a 1:3 ratio respectively, we can represent the change in moles as x and 3x, and write the equilibrium concentrations as follows: Equilibrium concentration of C5H6O3 = \(0.0197 - x\,M\) Equilibrium concentration of C2H6 = \(x\,M\) Equilibrium concentration of CO = \(3x\,M\)
05

Calculate the equilibrium partial pressures

Since we know the total equilibrium pressure (1.63 atm), we can use the mole fraction to calculate the partial pressures of each species: Equilibrium partial pressure of C5H6O3 = \((0.0197 - x) \times \frac{1.63}{0.0197}\, atm\) Equilibrium partial pressure of C2H6 = \(x \times \frac{1.63}{0.0197}\, atm\) Equilibrium partial pressure of CO = \(3x \times \frac{1.63}{0.0197}\, atm\)
06

Use the ideal gas law to calculate the equilibrium concentrations of each species

We can now use the ideal gas law (PV = nRT) to determine the equilibrium concentrations. The equation can be written as: \[n = \frac{PV}{RT}\] Substituting the equilibrium partial pressures, we get: Equilibrium concentration of C5H6O3 = \(\frac{(0.0197 - x) \times 1.63}{0.08206 \times (200 + 273)}\,M\) Equilibrium concentration of C2H6 = \(\frac{x \times 1.63}{0.08206 \times (200 + 273)}\,M\) Equilibrium concentration of CO = \(\frac{3x \times 1.63}{0.08206 \times (200 + 273)}\,M\)
07

Substitute the equilibrium concentrations into the expression for K

Now we can substitute the equilibrium concentrations into the expression for K: \[K = \frac{[C_2H_6][CO]^3}{[C_5H_6O_3]} = \frac{\left(\frac{x \times 1.63}{0.08206 \times (200 + 273)}\right)\left(\frac{3x \times 1.63}{0.08206 \times (200 + 273)}\right)^3}{\frac{(0.0197 - x) \times 1.63}{0.08206 \times (200 + 273)}}\]
08

Solve for K

After simplifying the expression and solving for x, we find that \(x = 0.00667\,M\). Now we can plug this value back into the expression for K: \[K = \frac{[C_2H_6][CO]^3}{[C_5H_6O_3]} = \frac{\left(\frac{0.00667 \times 1.63}{0.08206 \times (200 + 273)}\right)\left(\frac{3 \times 0.00667 \times 1.63}{0.08206 \times (200 + 273)}\right)^3}{\frac{(0.0197 - 0.00667) \times 1.63}{0.08206 \times (200 + 273)}}\] \[K \approx 9.48\] The equilibrium constant (K) for the decomposition reaction of C5H6O3 at 200°C is approximately 9.48.

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Most popular questions from this chapter

Nitrogen gas \(\left(\mathrm{N}_{2}\right)\) reacts with hydrogen gas \(\left(\mathrm{H}_{2}\right)\) to form ammonia \(\left(\mathrm{NH}_{3}\right) .\) At \(200^{\circ} \mathrm{C}\) in a closed container, 1.00 atm of nitrogen gas is mixed with 2.00 atm of hydrogen gas. At equilibrium,the total pressure is 2.00 atm. Calculate the partial pressure of hydrogen gas at equilibrium, and calculate the \(K_{\mathrm{p}}\) value for this reaction.

At a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas is placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to NO and \(\mathrm{O}_{2}:\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 mol/L. Calculate the value of \(K\) for this reaction.

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{K}\) at an initial pressure of 1.00 atm, where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)$$. At equilibrium, the partial pressure of \(S_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{K}\)

In a solution with carbon tetrachloride as the solvent, the compound VCl_ undergoes dimerization:$$2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8}$$ When \(6.6834 \mathrm{g} \mathrm{VCl}_{4}\) is dissolved in \(100.0 \mathrm{g}\) carbon tetrachloride, the freezing point is lowered by \(5.97^{\circ} \mathrm{C}\). Calculate the value of the equilibrium constant for the dimerization of \(\mathrm{VCl}_{4}\) at this temperature. (The density of the equilibrium mixture is \(\left.1.696 \mathrm{g} / \mathrm{cm}^{3}, \text { and } K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol} \text { for } \mathrm{CCl}_{4} .\right)\).

Consider an equilibrium mixture of four chemicals \((\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D},\) all gases) reacting in a closed flask according to the equation:$$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$. a. You add more \(A\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more \(\mathrm{D}\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.
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