An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and \(\mathrm{HI}\) at \(458^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{mol} \mathrm{H}_{2}, 0.112 \mathrm{mol} \mathrm{I}_{2},\) and 0.775 \(\mathrm{mol}\) HI in a 5.00 -L. vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 \(\mathrm{mol}\) of \(\mathrm{HI}\) ?

Dividing the initial moles of each substance by the volume of the container (5.00 L) will give us the initial concentrations in M (molarity): Initial Molarity of H2 = \(\frac{0.112 \ mol}{5.00 \ L}\) = 0.0224 M Initial Molarity of I2 = \(\frac{0.112 \ mol}{5.00 \ L}\) = 0.0224 M Initial Molarity of HI = \(\frac{0.775 \ mol}{5.00 \ L}\) = 0.155 M

Now, we need to adjust the initial concentrations by adding the additional amount of HI: New concentration of HI = \(\frac{(0.775 + 0.200) \ mol}{5.00 \ L}\) = \(\frac{0.975 \ mol}{5.00 \ L}\) = 0.195 M

The reaction is $$H_2 + I_2 \rightleftharpoons 2HI$$ To determine the equilibrium concentrations, we will set up a table showing the initial concentrations (I), changes (C), and equilibrium concentrations (E) for the reactants and products: Substance: H2 I2 2HI Initial (I): 0.0224 M 0.0224 M 0.195 M Change (C): -x -x +2x Equilibrium (E): 0.0224-x 0.0224-x 0.195+2x

The equilibrium constant expression for the reaction is: \(K_c = \frac{[HI]^2}{[H_2][I_2]}\) Since we are not given the value of Kc, we cannot directly solve the equation. However, we know that the initial conditions were at equilibrium, so we can use the initial concentrations to find Kc: Kc =\(\frac{[0.155]^2}{[0.0224][0.0224]}\) Kc = \(\frac{0.024025}{0.00050176}\) Kc = 47.9 Now that we have the value of Kc, we can use the equilibrium concentrations to find x: \(47.9 = \frac{[0.195+2x]^2}{[(0.0224-x)(0.0224-x)]}\)

To solve for x, we now have to solve the above equation. In practice, it might be necessary to use a numerical method, such as the quadratic formula, to find a solution. However, in this case, we can approximate the solution by assuming that x << 0.0224, making the quadratic equation solvable: \(47.9 = \frac{[0.195+2x]^2}{[(0.0224)(0.0224)]}\) Solving for x: x ≈ 0.00926 M

We now have the value of x, which we can use to calculate the equilibrium concentrations: Equilibrium concentration of H2 = 0.0224 - x = 0.01314 M Equilibrium concentration of I2 = 0.0224 - x = 0.01314 M Equilibrium concentration of HI = 0.195 + 2x = 0.21352 M Finally, we can find the equilibrium partial pressures using the ideal gas law: Equilibrium partial pressure of H2 = 0.01314 M \(\times\) 0.08206 L atm/mol K \(\times\) (458 + 273) K = 0.56 atm Equilibrium partial pressure of I2 = 0.01314 M \(\times\) 0.08206 L atm/mol K \(\times\) (458 + 273) K = 0.56 atm Equilibrium partial pressure of HI = 0.21352 M \(\times\) 0.08206 L atm/mol K \(\times\) (458 + 273) K = 9.09 atm Therefore, after the addition of 0.200 mol HI, the equilibrium partial pressures in the reaction mixture are 0.56 atm for both H2 and I2, and 9.09 atm for HI.