Question

When 2.00 mol of SO2Cl2 is placed in a 2.00-Lflask at 303 K, 56% of the SO2Cl2 decomposes to SO2 and Cl2:

SO2Cl2(g)⇌SO2(g)+Cl2(g)

K_c= 0.71 K_p= 18

Use the quilibrium constant you calculated above to determine the percentage of S0₂Cl₂ that decomposes when 2.00 mol of SO₂Cl₂ is placed in a 19.00 L vessel at 303K.

I am having a hard time determining the answer to this last part, please help! Thank you.

Answer 1

1) SO2Cl2(g) ⇌ SO2(g) + Cl2(g)

Initial 2 mol/2L 0 0

= 1 M

At equilibrium 1- 0.56 0.56 0.56

= 0.44

Hence,

Kc = 0.56 x 0.56/0.44 = 0.71

Kp = Kc (RT) ⁿg = 0.71 x (0.0821 x 303 ) ^(1+1-1) = 18

Therefore,

Kc= 0.71 , Kp = 18

2)   

SO2Cl2(g) ⇌ SO2(g) + Cl2(g)

Initial 2 mol/19 L 0 0

= 0.105 M

At equilibrium 0.105 -x x x

Kc= x.x /0.105-x

0.71 = x²/0.105-x

x² + 0.71 x - 0.07455 = 0

On solving,

x = 0.0928 M

Hence,

percentage of SO2Cl2 decomposed = 0.0928/0.105 x 100 = 88.3 %