Calculate Q: Q = [HBr]²/([H₂][Br₂]) = (0.300)²/(0.200 × 0.100) = 0.09/0.02 = 4.5. Since Q < Kc (55.3), reaction proceeds forward. Let x = mol/L of H₂ and Br₂ consumed. At equilibrium: [H₂] = 0.200 - x, [Br₂] = 0.100 - x, [HBr] = 0.300 + 2x. Substituting: Kc = (0.300 + 2x)²/[(0.200 - x)(0.100 - x)] = 55.3. Expanding: (0.09 + 1.2x + 4x²)/[(0.02 - 0.3x + x²)] = 55.3. Solving the resulting quadratic gives x = 0.0435 M, so [Br₂] = 0.100 - 0.0435 = 0.0565 M.

Because $$K_p$$ is very large, the reaction proceeds nearly to completion. Assume it goes to completion first: $$P_{H_2}$$ and $$P_{Cl_2}$$ become 0 atm, and $$P_{HCl}$$ becomes 2.0 atm. Then, let the reaction shift back to equilibrium by an amount x. At equilibrium, $$P_{H_2} = x$$, $$P_{Cl_2} = x$$, and $$P_{HCl} = 2.0 - 2x$$. The equilibrium expression is $$K_p = \frac{(P_{HCl})^2}{(P_{H_2})(P_{Cl_2})}$$. So, $$2.5 \times 10^{4} = \frac{(2.0 - 2x)^2}{x^2}$$. We can approximate $$2.0 - 2x \approx 2.0$$. This gives $$2.5 \times 10^{4} \approx \frac{(2.0)^2}{x^2}$$. Solving for x gives $$x^2 = \frac{4.0}{2.5 \times 10^4} = 1.6 \times 10^{-4}$$, so $$x \approx 0.013$$ atm. This is the equilibrium pressure of $$H_2$$.

First, calculate the reaction quotient, $$Q_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.20)(0.20)}{(0.20)} = 0.20$$. Since $$Q_c (0.20) > K_c (0.040)$$, the reaction will shift to the left to reach equilibrium. Let x be the change in concentration. At equilibrium, $$[PCl_5] = 0.20 + x$$, $$[PCl_3] = 0.20 - x$$, and $$[Cl_2] = 0.20 - x$$. Then, $$0.040 = \frac{(0.20-x)^2}{0.20+x}$$. This expands to the quadratic equation $$x^2 - 0.44x + 0.032 = 0$$. Solving gives $$x \approx 0.092 M$$. The equilibrium concentration of $$Cl_2$$ is $$0.20 - x = 0.20 - 0.092 = 0.108 M$$, which is approximately $$0.11 M$$.

From the stoichiometry of the reaction, for every 1 mole of $$H_2$$ formed, 2 moles of HI must have reacted. If $$[H_2]{eq} = 0.20 M$$, then the change in HI concentration is $$2 \times 0.20 M = 0.40 M$$. The initial concentration of HI was 2.0 M. Therefore, the equilibrium concentration of HI is the initial concentration minus the amount that reacted: $$[HI]{eq} = 2.0 M - 0.40 M = 1.6 M$$.

First, calculate the initial concentration of A: $$[A]_{initial} = \frac{4.0 \text{ mol}}{2.0 \text{ L}} = 2.0 M$$. Let x be the equilibrium concentration of B. Then $$[C] = x$$ and $$[A] = 2.0 - 2x$$. The equilibrium expression is $$K_c = \frac{[B][C]}{[A]^2} = \frac{(x)(x)}{(2.0 - 2x)^2} = 0.25$$. Taking the square root of both sides gives $$\frac{x}{2.0 - 2x} = \sqrt{0.25} = 0.50$$. Solving for x: $$x = 0.50(2.0 - 2x) = 1.0 - x$$. This gives $$2x = 1.0$$, so $$x = 0.50 M$$. The equilibrium concentration of B is $$0.50 M$$.

Since only product is present initially, the reaction will proceed in reverse. Let x be the equilibrium partial pressure of $$N_2O_4$$. The change in $$NO_2$$ pressure will be -2x. At equilibrium, $$P_{N_2O_4} = x$$ and $$P_{NO_2} = 1.0 - 2x$$. The equilibrium expression is $$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}$$. So, $$0.66 = \frac{(1.0 - 2x)^2}{x}$$. This gives the quadratic equation $$4x^2 - 4.66x + 1.0 = 0$$. Solving for x gives two possible values, but only $$x \approx 0.28$$ atm results in a positive pressure for $$NO_2$$. Thus, the equilibrium pressure of $$N_2O_4$$ is $$0.28$$ atm.

Let x be the change in pressure of $$PCl_5$$. At equilibrium, $$P_{PCl_5} = 2.0 - x$$, $$P_{PCl_3} = x$$, and $$P_{Cl_2} = x$$. The equilibrium expression is $$K_p = \frac{(P_{PCl_3})(P_{Cl_2})}{P_{PCl_5}}$$. So, $$1.0 = \frac{x^2}{2.0 - x}$$. This rearranges to the quadratic equation $$x^2 + x - 2.0 = 0$$, which factors to $$(x+2)(x-1)=0$$. The only positive root is $$x=1.0$$ atm. The equilibrium partial pressures are $$P_{PCl_5} = 1.0$$ atm, $$P_{PCl_3} = 1.0$$ atm, and $$P_{Cl_2} = 1.0$$ atm. The total pressure is the sum: $$1.0 + 1.0 + 1.0 = 3.0$$ atm.

First, calculate the initial pressure of $$SO_2Cl_2$$ using the ideal gas law: $$P = \frac{nRT}{V} = \frac{(0.10 \text{ mol})(0.08206 \text{ L atm/mol K})(375 \text{ K})}{1.0 \text{ L}} \approx 3.08$$ atm. Let x be the change in pressure. At equilibrium, $$P_{SO_2Cl_2} = 3.08 - x$$, and $$P_{SO_2} = P_{Cl_2} = x$$. $$K_p = \frac{(P_{SO_2})(P_{Cl_2})}{P_{SO_2Cl_2}} = \frac{x^2}{3.08 - x} = 2.9$$. Rearranging yields the quadratic equation $$x^2 + 2.9x - 8.93 = 0$$. The positive root is $$x \approx 1.87$$ atm. Therefore, the equilibrium partial pressure of $$SO_2$$ is approximately 1.9 atm.

Mixing equal volumes halves the initial concentrations to 0.10 M for both $$H^+$$ and $$C_2H_3O_2^-$$. Since K is very large, the reaction proceeds almost to completion, forming 0.10 M $$HC_2H_3O_2$$ and leaving negligible amounts of reactants. To find the small amount of $$H^+$$ left, assume the reaction goes to completion and then shifts back. Let $$[H^+]_{eq} = x$$. At equilibrium, $$[H^+] = [C_2H_3O_2^-] = x$$ and $$[HC_2H_3O_2] = 0.10 - x \approx 0.10$$. Then $$K = \frac{[HC_2H_3O_2]}{[H^+][C_2H_3O_2^-]} \approx \frac{0.10}{x^2} = 5.6 \times 10^4$$. Solving for x gives $$x^2 = \frac{0.10}{5.6 \times 10^4} \approx 1.78 \times 10^{-6}$$, so $$x = [H^+] \approx 1.3 \times 10^{-3} M$$.

Let x be the change in concentration of $$I_2$$. At equilibrium, $$[I_2] = 0.050 - x$$ and $$[I] = 2x$$. The equilibrium expression is $$K_c = \frac{[I]^2}{[I_2]}$$. So, $$3.8 \times 10^{-5} = \frac{(2x)^2}{0.050 - x}$$. Since $$K_c$$ is small, we can approximate $$0.050 - x \approx 0.050$$. The equation becomes $$3.8 \times 10^{-5} \approx \frac{4x^2}{0.050}$$. Solving for $$x^2$$ gives $$x^2 = \frac{(3.8 \times 10^{-5})(0.050)}{4} = 4.75 \times 10^{-7}$$. So, $$x \approx 6.9 \times 10^{-4} M$$. The equilibrium concentration of I is $$2x$$, which is $$2 \times(6.9 \times 10^{-4}) \approx 1.38 \times 10^{-3} M$$, or approximately $$1.4 \times 10^{-3} M$$.