Consider the following reaction at \(725^{\circ} \mathrm{C} :\) $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ At equilibrium, a \(4.50-\mathrm{L}\) container has 2.6 \(\mathrm{g}\) of carbon, \(\mathrm{CO}_{2}\) at a partial pressure of \(0.0020 \mathrm{atm},\) and a total pressure of 0.572 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(725^{\circ} \mathrm{C}\)

First, we need to convert the given mass of carbon (2.6 g) into moles using its molar mass (12.01 g/mol): \[ \text{moles of carbon} = \frac{2.6\,\text{g}}{12.01\,\text{g/mol}} \approx 0.2164\,\text{mol} \]

Using the balanced chemical equation and the stoichiometry, we know that: \[ \text{moles of CO} = 2 \times \text{moles of carbon} = 2 \times 0.2164\,\text{mol} = 0.4328\,\text{mol} \] Since the partial pressure of CO\(_2\) is given as 0.0020 atm, we can calculate the moles of CO\(_2\) using the ideal gas law, knowing the temperature and volume. Rearranging the ideal gas law, \(PV=nRT\), for moles (n): \[ n = \frac{PV}{RT} \] Given that \(R = 0.0821\,\frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}}\) and the temperature \(T = 725 + 273.15 = 998.15\,\text{K}\): \[ \text{moles of CO}_{2} = \frac{(0.0020\,\text{atm})(4.50\,\text{L})}{(0.0821\,\frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}})(998.15\,\text{K})} \approx 1.22 \times 10^{-4}\,\text{mol} \]

We know that the total pressure is 0.572 atm and the partial pressure of CO\(_2\) is 0.0020 atm, thus, the partial pressure of CO can be found by subtracting the CO\(_2\) partial pressure from the total pressure: \[ \text{partial pressure of CO} = 0.572\,\text{atm} - 0.0020\,\text{atm} = 0.570\,\text{atm} \]

Now that we have the partial pressures of all the species involved in the balanced chemical reaction, we can calculate the equilibrium constant \(K_p\): \[ K_p = \frac{[\text{CO}]^2}{[\text{C}][\text{CO}_2]} \] Since Carbon is in solid form, its concentration does not affect the equilibrium constant, and the expression becomes: \[ K_p = \frac{[\text{CO}]^2}{[\text{CO}_2]} \] Plug in the partial pressures: \[ K_p = \frac{(0.570\,\text{atm})^2}{(0.0020\,\text{atm})} \approx 163,225 \] Thus, the equilibrium constant \(K_p\) for this reaction at 725°C is approximately 163,225.