Questions on Chemical Equilibrium

[candiishop]

The decomposition of sulfuryl chloride, SO2Cl2, is described by the following chemical equation:

 

SO2Cl2(g) <-----> SO2(g) + Cl2(g)

 

(i) At 670 K, the following equilibrium concentrations were observed:

[sO2Cl2] = 0.0678 M, [sO2] = 0.0243 M, and [Cl2] = 0.121 M. Calculate the equilibrium constant Kc for the reaction at 670 K.

 

Is this right? Kc = [sO2] [Cl2]

= 0.0243 x 0.121

= 0.00294

 

(ii) From your answer in (i), which side of the reaction does the equilibrium favour at 670 K? Justify your choice.

 

(iii) At the higher temperature of 1250 K, the value of Kc is 85.3.

If 0.152 moles of SO2Cl2(g), 0.336 moles of SO2(g), and 2.34 moles of Cl2(g)

were placed in a 2.50 L flask at 1250 K, would the amount of SO2 increase or

decrease as the reaction proceeded to equilibrium? Show your working.

 

SO2Cl2(g) <-----> SO2(g) + Cl2(g)

 

85.3 = [x] [x]

[x]^2 = 85.3

[x] = square root of 85.3

:. [x] = 9.236 M

[Mendelejev]

An equilibrium constant of a reaction A + B => C + D is

 

K = [C]eq[D]eq / [A]eqeq

 

So, in your case, it will be [sO2]eq [Cl2]eq / [sO2Cl2]eq

 

If K > 10³ the reaction will be sponteneous and you will form C and D

If K < 10-³ the reaction will not occur. C + D would react sponeneously to A + B, but not in the opposite direction

If 10-³ < K < 10³ you have an equilibrium. Some of your ammount will convert to C + D but you will get an equilibrium with A, B, C, and D

 

"At the higher temperature of 1250 K, the value of Kc is 85.3.

If 0.152 moles of SO2Cl2(g), 0.336 moles of SO2(g), and 2.34 moles of Cl2(g)

were placed in a 2.50 L flask at 1250 K, would the amount of SO2 increase or

decrease as the reaction proceeded to equilibrium? Show your working."

 

I think : Calculate the concentrations of SO2Cl2, SO2, Cl2

Calculate the reaction constant [sO2][Cl2] / [sO2CL2]

and compare it with Kc = 85,3.

 

If K < Kc SO2Cl2 will convert to SO2 and Cl2 (NOT COMPLETELY !!!)

If K = Kc you have the equilibrium

If K > Kc their is to much SO2 and Cl2 and they will react and form SO2Cl2 (again, not completely)

 

So, I hope, what I just said was right ! Good luck

[candiishop]

Many thanks for reply although i'm still unsure how to do the question or my answers are correct.