Consider the decomposition of the compound $\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}$ as follows: $$\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g)$$ 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-\mathrm{L}$ flask and heated to \(200 .^{\circ} \mathrm{C},\) the pres- sure in the flask gradually rose to 1.63 \(\mathrm{atm}\) and remained at that value. Calculate \(K\) for this reaction.

First, we need to find the initial moles of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) using the given mass and the molar mass of the compound. Molar mass of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3} = 5\times 12.01 + 6\times 1.01 + 3\times 16.00 = 114.11 \ \text{g/mol}\). Initial moles \((n_i)\) of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3} = \frac{5.63\ \text{g}}{114.11\ \text{g/mol}}\)

Set up an ICE (initial, change, equilibrium) table to record the changes in moles of reactants and products during the reaction. Reaction: $$\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g)$$ | | \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) | \(\mathrm{C}_{2} \mathrm{H}_{6}\) | \(3\ \mathrm{CO}\) | |---------- |------------------------------------ |---------------------------- |-------------- | | Initial | \(n_i\) | 0 | 0 | | Change | \(-x\) | \(+x\) | \(+3x\) | | Equilibrium | \(n_i - x\) | \(x\) | \(3x\) |

Using the final pressure \((P_f = 1.63\ \text{atm})\) and ideal gas law, we can find the total moles of gas at equilibrium and use it to calculate \(x\). Ideal Gas Law: \(PV=nRT\) Total moles of gas \((n_t)\) at equilibrium: $$n_t = \frac{PV}{RT} = \frac{(1.63\ \text{atm})(2.50\ \text{L})}{(0.0821\ \text{L.atm/mol.K})(293\ \text{K})}$$ As the total moles of gas at equilibrium are equal to the sum of moles of all the gases, we can write, $$n_t = n_{\text{C}_{5} \text{H}_{6} \text{O}_{3}} + n_{\text{C}_{2} \text{H}_{6}} + n_{\text{CO}} = (n_i - x) + x + 3x$$ Now we can substitute the known values and calculate \(x\).

Once we have \(x\), we can find the equilibrium concentrations of \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\), \(\mathrm{C}_{2} \mathrm{H}_{6}\), and \(\mathrm{CO}\) using the equilibrium moles and the given volume of the flask. Equilibrium concentration (\(\textit{conc}\)) for each species: \([\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}]_{eq} = \frac{n_{\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}}}{V} = \frac{n_i - x}{2.50\ \text{L}}\) \([\mathrm{C}_{2} \mathrm{H}_{6}]_{eq} = \frac{n_{\mathrm{C}_{2} \mathrm{H}_{6}}}{V} = \frac{x}{2.50\ \text{L}}\) \([\mathrm{CO}]_{eq} = \frac{n_{\mathrm{CO}}}{V} = \frac{3x}{2.50\ \text{L}}\)

Finally, we can calculate the equilibrium constant using the known equilibrium concentrations and the balanced chemical equation: $$K = \frac{[\mathrm{C}_{2} \mathrm{H}_{6}]_{eq} [\mathrm{CO}]_{eq}^3}{[\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}]_{eq}}$$ Now, substitute the values for the equilibrium concentrations and calculate the value of \(K\).