Consider the reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) Kp=2.91×103 at 298 K In a reaction at equilibrium, the partial pressure...

Consider the reaction:
SO2Cl2(g)⇌SO2(g)+Cl2(g) Kp=2.91×10³ at 298 K
In a reaction at equilibrium, the partial pressure of SO2 is 0.135 atm and that of Cl2 is 0.342 atm .

Part A

What is the partial pressure of SO2Cl2 in this mixture?

Solutions

Expert Solution

Solution :-

Lets make the ICE table

SO2Cl2(g) ⇌ SO2(g) + Cl2(g)

0 0.135 0.342

+x -x -x

X 0.135-x 0.342-x

Kp = [SO2][Cl2]/[SO2Cl2]

2.91*10^3 = [0.135-x][0.342-x]/[x]

2.91*10^3 *x = [0.135-x][0.342-x]

Solving for x we get

X= 1.58*1^-5

Therefore the partial pressure of the SO2Cl2 at equilibrium = 1.58*10^-5 atm

queen_honey_blossom answered 4 weeks ago

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