Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the reaction: PCl5 (g) ⇔ PCl3 (g) + Cl2 (g). At 250°C, 0.250 M PCl5 is added to a flask. If KC= 1.80, what are the equilibrium concentrations of each gas? A) [PCl5] = 1.25 M, [PCl3] = 0.474 M, [Cl2] = 0.474 M B) [PCl5] = 2.27 M, [PCl3] = 2.02 M, [Cl2] = 2.02 M C) [PCl5] = 0.0280 M, [PCl3] = 0.222 M, [Cl2] = 0.222 M D) [PCl5] = 1.80 M, [PCl3] = 1.80 M, [Cl2] = 1.80 M

Respuesta :

Answer:

C) [PCl5] = 0.0280 M, [PCl3] = 0.222 M, [Cl2] = 0.222 M

Explanation:

Moles of [PCl₅] = 0.250 M

Considering the ICE table for the equilibrium as:

                     PCl₅ (g)      ⇔          PCl₃ (g) +       Cl₂ (g)

t = o               0.250

t = eq                -x                             x                      x

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Moles at eq: 0.250-x                       x                      x        

The expression for the equilibrium constant is:

[tex]K_c=\frac {[PCl_3][Cl_2]}{[PCl_5]}=1.80[/tex]

So,

[tex]\frac{x^2}{0.250-x}=1.80[/tex]

x = 0.222 M

Equilibrium concentrations :

[PCl₃] = [Cl₂] = 0.222 M

[PCl₅] = 0.250 - 0.222 = 0.280 M

Option C is correct.