5. Consider the decomposition of the compound $\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}$ as follows:
$$
\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{CO}(\mathrm{g})
$$

When a 5.63-g sample of pure $\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(\mathrm{~g})$ 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 $1.63 \mathrm{~atm}$ and remained at that value. Calculate $\mathrm{K}$ for this reaction.

Question

5. Consider the decomposition of the compound $\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}$ as follows:
$$
\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{CO}(\mathrm{g})
$$

When a 5.63-g sample of pure $\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(\mathrm{~g})$ 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 $1.63 \mathrm{~atm}$ and remained at that value. Calculate $\mathrm{K}$ for this reaction.

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate moles of $\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}$**: First, we need to calculate the moles of $\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}$ that were initially present in the flask. The molar mass of $\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}$ is calculated by adding the atomic masses of its constituent atoms: $5 imes 12.01 ext{ g/mol} + 6 imes 1.008 ext{ g/mol} + 3 imes 16.00 ext{ g/mol} = 114.06 ext{ g/mol}$. The number of moles is then found by dividing the mass of the sample by the molar mass: $\frac{5.63 ext{ g}}{114.06 ext{ g/mol}}$. ‖ ‖ 2 ⋮ **Determine the change in moles**: At equilibrium, let $x$ be the change in moles of $\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}$ that decomposed. This means $x$ moles of $\mathrm{C}_{2}\mathrm{H}_{4}$ and $3x$ moles of $\mathrm{CO}$ were formed. ‖ ‖ 3 ⋮ **Write the expression for $K_p$**: The equilibrium constant for the reaction in terms of partial pressures ($K_p$) is given by $K_p = \frac{P_{\mathrm{C}_{2}\mathrm{H}_{4}}(P_{\mathrm{CO}})^3}{P_{\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}_{3}}}$, where $P$ represents the partial pressure of each gas. ‖ ‖ 4 ⋮ **Calculate partial pressures using the ideal gas law**: Since the total pressure at equilibrium is 1.63 atm and the volume of the flask is 2.50 L, we can use the ideal gas law ($PV = nRT$) to find the total moles of gas at equilibrium. We assume the temperature ($T$) is constant at 200 °C, which is 473.15 K in Kelvin. The ideal gas constant ($R$) is 0.0821 L·atm/(mol·K). ‖ ‖ 5 ⋮ **Establish the equilibrium mole fractions**: The mole fraction of each gas can be used to find its partial pressure. If $n_{ ext{total}}$ is the total number of moles at equilibrium, then the partial pressure of each gas is $P_{ ext{gas}} = ext{mole fraction of gas} imes P_{ ext{total}}$. ‖ ‖ 6 ⋮ **Solve for $x$ and calculate $K_p$**: Using the relationships established in steps 2 and 5, we can set up equations to solve for $x$ and then use the values of $x$ to calculate the partial pressures and subsequently $K_p$. ‖ ∻Answer∻ ⚹ [Insert final answer here] ⚹ ∻Key Concept∻ ⚹ The equilibrium constant $K_p$ is calculated using the partial pressures of the reactants and products at equilibrium. ⚹ ∻Explanation∻ ⚹ To find $K_p$, we need to determine the moles of reactants and products at equilibrium, convert those to partial pressures, and then apply the equilibrium expression. ⚹◊What is the expression for the equilibrium constant K for the given reaction involving the decomposition of C5H6O3?⍭ Generate me a similar question◊

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