One mole of \(\mathrm{N}_{2}\) and 3 moles of \(\mathrm{H}_{2}\) are placed in a flask at \(397^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is found to be 0.21. The \(K_{p}\) for the reaction is \(4.31 \times 10^{-4}.\)

You need to understand the given information and set up your strategy. You are given that 1 mole of \(N_{2}\) and 3 moles of \(H_{2}\) are placed in a flask to react via the equation: \(N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \) . The reaction reaches equilibrium at 397°C, the mole fraction of NH3 is given as 0.21, and \(K_p = 4.31 \times 10^{-4} \).

The mole fraction of \(NH_3\) equals the partial pressure of \(NH_3\), \(x_{NH_3}\), over the total pressure \(P_{Total}\): \(x_{NH_3} = \frac{P_{NH_3}}{P_{Total}} \). We can rearrange this to calculate the partial pressure of \(NH_3\): \(P_{NH_3} = x_{NH_3} \times P_{Total} \). Using this, the fact that we need the total pressure, and the given \(K_p\) value, we can form the equation: \(K_p = \frac{P_{NH_3}^{2}}{P_{N_2} P_{H_2}^{3}} = \frac{(x_{NH_3} \times P_{Total})^{2}}{(0.79 \times P_{Total}) (0.79^{2} \times P_{Total}^{3})} = \frac{(0.21 \times P_{Total})^{2}}{(0.79 \times P_{Total}) (0.79^{3} \times P_{Total}^{3})} \). Note that the mole fractions of \(N_{2}\) and \(H_{2}\) are \(1-0.21 = 0.79 \) because at equilibrium the sum of all mole fractions equals 1.

Upon rearranging the equation from step 2, we get \(P_{Total} = \sqrt[5]{K_p} \), where the root is 5 because the denominator in step 2 has \(P_{Total}\) raised to the power 5. Substituting the given value for \(K_p\), we find \(P_{Total} = \sqrt[5]{4.31 \times 10^{-4}}\).

Calculate \(P_{Total}\) to get the final answer.