Nitric oxide and bromine at initial partial pressures of \(98.4\) and \(41.3\) torr, respectively, were allowed to react at \(300 . \mathrm{K}\). At equilibrium the total pressure was \(110.5\) torr. The reaction is $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ a. Calculate the value of \(K_{\mathrm{p}}\). b. What would be the partial pressures of all species if \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), both at an initial partial pressure of \(0.30 \mathrm{~atm}\), were allowed to come to equilibrium at this temperature?

First, let's determine the change in partial pressures for all species. We are given the following initial pressures: P(NO)initial = 98.4 torr P(Br2)initial = 41.3 torr P(NOBr)initial = 0 torr (since NOBr initially is not present) The total equilibrium pressure is given as 110.5 torr. Let x represent the change in partial pressures: P(NO) = 98.4 - 2x torr P(Br2) = 41.3 - x torr P(NOBr) = 2x torr Now we can write the equilibrium expression: \( K_p = \frac{P(NOBr)^2}{P(NO)^2 \times P(Br_2)} \) Substitute the expressions in terms of x: \( K_p = \frac{(2x)^2}{(98.4-2x)^2 \times (41.3-x)} \) Since we have the total equilibrium pressure, we can write the equation: Total pressure = P(NO) + P(Br2) + P(NOBr) 110.5 = (98.4 - 2x) + (41.3 - x) + (2x) Solving for x: 110.5 = 98.4 + 41.3 - x x = 29.2 torr Now, substitute the value of x back into the expressions for the partial pressures at equilibrium: P(NO) = 98.4 - 2(29.2) = 40 torr P(Br2) = 41.3 - 29.2 = 12.1 torr P(NOBr) = 2(29.2) = 58.4 torr Plug the equilibrium partial pressures back into the equilibrium expression: \( K_p = \frac{58.4^2}{40^2 \times 12.1} \) Now, calculate the value of Kp: \( K_p \approx 0.412 \)

Given the initial partial pressures of NO and Br2 (both at 0.30 atm), we need to find the equilibrium partial pressures for all species. First, convert the initial pressures to torr: P(NO)initial = 0.30 atm × (760 torr/atm) = 228 torr P(Br2)initial = 0.30 atm × (760 torr/atm) = 228 torr Let y represent the change in partial pressures: P(NO) = 228 - 2y torr P(Br2) = 228 - y torr P(NOBr) = 2y torr Using the previously calculated Kp value (0.412), we can write the equilibrium expression: \( 0.412 = \frac{(2y)^2}{(228-2y)^2 \times (228-y)} \) Solving this equation for y is a bit more challenging, so it's recommended to use a numerical solving method such as the Newton-Raphson method or trial and error. After solving for y, we get: y ≈ 185.21 torr Now, substitute the value of y back into the expressions for the partial pressures and convert them back to atm: P(NO) = (228 - 2(185.21)) / 760 ≈ 0.012 atm P(Br2) = (228 - 185.21) / 760 ≈ 0.056 atm P(NOBr) = 2(185.21) / 760 ≈ 0.49 atm So, at equilibrium, the partial pressures of the species are: P(NO) ≈ 0.012 atm P(Br2) ≈ 0.056 atm P(NOBr) ≈ 0.49 atm