a) La concentración de C l X 2 ( g ) \ce{Cl2(g)} Cl X 2 ( g ) en el equilibrio. La temperatura en Kelvin es T = 250 + 273.15 = 523.15 K T = 250 + 273.15 = 523.15 \text{ K} T = 250 + 273.15 = 523.15 K . Las concentraciones iniciales de los gases son:
[ P C l X 5 ] 0 = 2.0 mol 5 L = 0.40 M [\ce{PCl5}]_0 = \frac{2.0 \text{ mol}}{5 \text{ L}} = 0.40 \text{ M} [ PCl X 5 ] 0 = 5 L 2.0 mol = 0.40 M [ P C l X 3 ] 0 = 1.0 mol 5 L = 0.20 M [\ce{PCl3}]_0 = \frac{1.0 \text{ mol}}{5 \text{ L}} = 0.20 \text{ M} [ PCl X 3 ] 0 = 5 L 1.0 mol = 0.20 M [ C l X 2 ] 0 = 0 M [\ce{Cl2}]_0 = 0 \text{ M} [ Cl X 2 ] 0 = 0 M Se establece la tabla ICE (Inicial, Cambio, Equilibrio) para las concentraciones:
Especie Inicial (M) Cambio (M) Equilibrio (M) P C l X 5 0.40 − x 0.40 − x P C l X 3 0.20 + x 0.20 + x C l X 2 0 + x x \begin{array}{|l|c|c|c|} \hline \text{Especie} & \text{Inicial (M)} & \text{Cambio (M)} & \text{Equilibrio (M)} \\ \hline \ce{PCl5} & 0.40 & -x & 0.40 - x \\ \ce{PCl3} & 0.20 & +x & 0.20 + x \\ \ce{Cl2} & 0 & +x & x \\ \hline \end{array} Especie PCl X 5 PCl X 3 Cl X 2 Inicial (M) 0.40 0.20 0 Cambio (M) − x + x + x Equilibrio (M) 0.40 − x 0.20 + x x La expresión de la constante de equilibrio K c K_c K c es:
K c = [ P C l X 3 ] [ C l X 2 ] [ P C l X 5 ] K_c = \frac{[\ce{PCl3}] [\ce{Cl2}]}{[\ce{PCl5}]} K c = [ PCl X 5 ] [ PCl X 3 ] [ Cl X 2 ] Sustituyendo los valores en el equilibrio y el valor de K c = 0.042 K_c = 0.042 K c = 0.042 :
0.042 = ( 0.20 + x ) ( x ) ( 0.40 − x ) 0.042 = \frac{(0.20 + x)(x)}{(0.40 - x)} 0.042 = ( 0.40 − x ) ( 0.20 + x ) ( x ) Reordenando la ecuación, se obtiene una ecuación cuadrática:
0.042 ( 0.40 − x ) = 0.20 x + x 2 0.042(0.40 - x) = 0.20x + x^2 0.042 ( 0.40 − x ) = 0.20 x + x 2 0.0168 − 0.042 x = 0.20 x + x 2 0.0168 - 0.042x = 0.20x + x^2 0.0168 − 0.042 x = 0.20 x + x 2 x 2 + 0.242 x − 0.0168 = 0 x^2 + 0.242x - 0.0168 = 0 x 2 + 0.242 x − 0.0168 = 0 Resolviendo la ecuación cuadrática a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 con x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c :
x = − 0.242 ± ( 0.242 ) 2 − 4 ( 1 ) ( − 0.0168 ) 2 ( 1 ) x = \frac{-0.242 \pm \sqrt{(0.242)^2 - 4(1)(-0.0168)}}{2(1)} x = 2 ( 1 ) − 0.242 ± ( 0.242 ) 2 − 4 ( 1 ) ( − 0.0168 ) x = − 0.242 ± 0.058564 + 0.0672 2 x = \frac{-0.242 \pm \sqrt{0.058564 + 0.0672}}{2} x = 2 − 0.242 ± 0.058564 + 0.0672 x = − 0.242 ± 0.125764 2 x = \frac{-0.242 \pm \sqrt{0.125764}}{2} x = 2 − 0.242 ± 0.125764 x = − 0.242 ± 0.35463 2 x = \frac{-0.242 \pm 0.35463}{2} x = 2 − 0.242 ± 0.35463 Se obtienen dos posibles valores para x x x : x 1 = 0.056315 x_1 = 0.056315 x 1 = 0.056315 y x 2 = − 0.298315 x_2 = -0.298315 x 2 = − 0.298315 . Dado que la concentración no puede ser negativa, se toma el valor positivo.
x = 0.056315 M x = 0.056315 \text{ M} x = 0.056315 M La concentración de C l X 2 ( g ) \ce{Cl2(g)} Cl X 2 ( g ) en el equilibrio es:
[ C l X 2 ] eq = x ≈ 0.056 M [\ce{Cl2}]_{\text{eq}} = x \approx 0.056 \text{ M} [ Cl X 2 ] eq = x ≈ 0.056 M b) El valor de K p K_p K p a esa misma temperatura y la presión en el recipiente una vez alcanzado el equilibrio. Para calcular K p K_p K p , se utiliza la relación K p = K c ( R T ) Δ n K_p = K_c (RT)^{\Delta n} K p = K c ( R T ) Δ n . El cambio en el número de moles de gas, Δ n \Delta n Δ n , es:
Δ n = ( 1 + 1 ) − 1 = 1 \Delta n = (1 + 1) - 1 = 1 Δ n = ( 1 + 1 ) − 1 = 1 Sustituyendo los valores de K c K_c K c , R R R , T T T y Δ n \Delta n Δ n :
K p = 0.042 × ( 0.082 atm ⋅ L ⋅ K − 1 ⋅ mol − 1 × 523.15 K ) 1 K_p = 0.042 \times (0.082 \text{ atm} \cdot \text{L} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \times 523.15 \text{ K})^1 K p = 0.042 × ( 0.082 atm ⋅ L ⋅ K − 1 ⋅ mol − 1 × 523.15 K ) 1 K p = 0.042 × 42.8983 ≈ 1.8 K_p = 0.042 \times 42.8983 \approx 1.8 K p = 0.042 × 42.8983 ≈ 1.8 Para calcular la presión total en el recipiente, primero se determinan las moles de cada especie en el equilibrio:
n eq ( C l X 2 ) = [ C l X 2 ] eq × V = 0.056315 M × 5 L = 0.281575 mol n_{\text{eq}}(\ce{Cl2}) = [\ce{Cl2}]_{\text{eq}} \times V = 0.056315 \text{ M} \times 5 \text{ L} = 0.281575 \text{ mol} n eq ( Cl X 2 ) = [ Cl X 2 ] eq × V = 0.056315 M × 5 L = 0.281575 mol n eq ( P C l X 5 ) = ( 0.40 − x ) × V = ( 0.40 − 0.056315 ) M × 5 L = 0.343685 M × 5 L = 1.718425 mol n_{\text{eq}}(\ce{PCl5}) = (0.40 - x) \times V = (0.40 - 0.056315) \text{ M} \times 5 \text{ L} = 0.343685 \text{ M} \times 5 \text{ L} = 1.718425 \text{ mol} n eq ( PCl X 5 ) = ( 0.40 − x ) × V = ( 0.40 − 0.056315 ) M × 5 L = 0.343685 M × 5 L = 1.718425 mol n eq ( P C l X 3 ) = ( 0.20 + x ) × V = ( 0.20 + 0.056315 ) M × 5 L = 0.256315 M × 5 L = 1.281575 mol n_{\text{eq}}(\ce{PCl3}) = (0.20 + x) \times V = (0.20 + 0.056315) \text{ M} \times 5 \text{ L} = 0.256315 \text{ M} \times 5 \text{ L} = 1.281575 \text{ mol} n eq ( PCl X 3 ) = ( 0.20 + x ) × V = ( 0.20 + 0.056315 ) M × 5 L = 0.256315 M × 5 L = 1.281575 mol El número total de moles en el equilibrio es la suma de las moles de cada especie:
n total = 0.281575 + 1.718425 + 1.281575 = 3.281575 mol n_{\text{total}} = 0.281575 + 1.718425 + 1.281575 = 3.281575 \text{ mol} n total = 0.281575 + 1.718425 + 1.281575 = 3.281575 mol Finalmente, se aplica la ecuación de los gases ideales ( P total V = n total R T P_{\text{total}} V = n_{\text{total}} RT P total V = n total R T ):
P total = n total R T V P_{\text{total}} = \frac{n_{\text{total}} RT}{V} P total = V n total R T P total = 3.281575 mol × 0.082 atm ⋅ L ⋅ K − 1 ⋅ mol − 1 × 523.15 K 5 L P_{\text{total}} = \frac{3.281575 \text{ mol} \times 0.082 \text{ atm} \cdot \text{L} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \times 523.15 \text{ K}}{5 \text{ L}} P total = 5 L 3.281575 mol × 0.082 atm ⋅ L ⋅ K − 1 ⋅ mol − 1 × 523.15 K P total = 141.018 atm ⋅ L 5 L ≈ 28.2 atm P_{\text{total}} = \frac{141.018 \text{ atm} \cdot \text{L}}{5 \text{ L}} \approx 28.2 \text{ atm} P total = 5 L 141.018 atm ⋅ L ≈ 28.2 atm 
