Question

The equilibrium constant, Kc, for the following reaction is 4.76×10-4 at 431 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently large sample of PCl5(g) is introduced into an evacuated vessel at 431 K, the equilibrium concentration of Cl2(g) is found to be 0.233 M. Calculate the concentration of PCl5 in the equilibrium mixture. M

Answer 1

Answer:

Explanation:

Step 1: Data given

The equilibrium constant, Kc, for the following reaction is 4.76 * 10^-4 at 431 K

The equilibrium concentration of Cl2(g) is  0.233 M

Step 2: The balanced equation

PCl5(g)  ⇄ PCl3(g) + Cl2(g)

Step 3: The initial concentration

[PCl5]= Y M

[PCl] = 0M

[Cl2] = 0M

Step 4: Calculate the concentration at equilibrium

[PCl5] = Y + X M = Y - 0.233 M

[PCl]= XM = 0.233 M

[Cl2]= XM = 0.233 M

Step 5: Define Kc

Kc =  [Cl2]* [PCl3] / [PCl5]

4.76 * 10^-4 = 0.233² / (Y -0.233)

0.000476 = 0.05429 / (Y - 0.233)

Y - 0.233 = 0.05429 / 0.000476

Y - 0.233 = 114.05 M

Y = 114.283 M = the initial concentration

The concentration of PCl5 at the equilibrium is 114.05 M