At a particular temperature, \(K_{\mathrm{p}}=1.00 \times 10^{2}\) for the reaction $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g)$$ If 2.00 atm of \(\mathrm{H}_{2}(g)\) and 2.00 atm of \(\mathrm{I}_{2}(g)\) are introduced into a \(1.00-\mathrm{L}\) container, calculate the equilibrium partial pressures of all species.

We are given the following initial information: - Reaction: \(H_2(g) + I_2(g) \leftrightharpoons 2 HI(g)\) - \(K_p = 1.00 \times 10^2\) - Initial partial pressure of \(H_2 = 2.00\, atm\) - Initial partial pressure of \(I_2 =2.00\, atm\) - Volume of the container = \(1.00\,L\)

An ICE table is a useful way to organize the information we have about how concentrations or pressures change during a chemical reaction. In this table, the columns represent the species involved in the reaction, and the rows represent the initial partial pressures (I), the change in partial pressures during the reaction (C), and the equilibrium partial pressures (E). The ICE table looks like this: | | H_2 | I_2 | 2 HI | |---|-----|-----|-----| | I | 2.00 atm | 2.00 atm | 0.00 atm | | C | -x | -x | +2x | | E | 2.00-x | 2.00-x | 2x | Here, "x" represents the decrease in the partial pressures of H_2 and I_2. Since the stoichiometry of the reaction shows that for every one mole of H_2 and one mole of I_2 consumed, two moles of HI are produced, we represent the change in HI with +2x.

Recall that the K_p is the ratio of the products to the reactants raised to their stoichiometric coefficients represented by partial pressures. For the given reaction: \(K_p = \frac{[HI]^2}{[H_2][I_2]}\) Substitute the equilibrium values from the ICE table: \(1.00x10^2 = \frac{(2x)^2}{(2-x)(2-x)}\)

Now we can solve the equation for x: \(100 = \frac{4x^2}{(2-x)^2}\) \(100(2-x)^2 = 4x^2\) \(100(4 - 4x + x^2) = 4x^2\) \(400 - 400x + 100x^2 = 4x^2\) \(96x^2 - 400x + 400 = 0\) This is a quadratic equation, which we can solve using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Where a = 96, b = -400, and c = 400. After solving for x, we get two possible values: \(x = 4.08\) and \(x = 1.08\). However, x cannot be greater than the initial partial pressures of H_2 and I_2 (2.00 atm), so we reject the value 4.08. Therefore, \(x = 1.08\). Finally, we can determine the equilibrium partial pressures for all species: - \(H_2\) and \(I_2\): \(2.00 - 1.08 = 0.92\, atm\) - \(HI\): \(2 \times 1.08 = 2.16\, atm\) The equilibrium partial pressures of the species are as follows: \(H_2: 0.92\, atm\) \(I_2: 0.92\, atm\) \(HI: 2.16\, atm\)