Question

The equilibrium reaction below has the Kc = 4.00. If the volume of the system at equilibrium is decreased from 4.00 liters to 2.00 liters, how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer.

Answer 1

Step-by-step explanation

Given

2 SO3 (g) ---> 2 SO2 (g) + O2 (g)

Kc = 4.00

initial volume = 4.00 L

Final volume = 2.00 L

Required: Calculate the value of the reaction quotient, Q

Solution

`Kc\text{ = }([O_2][SO_2]^2)/([SO_3]^2)\text{ = 4.00}`

PV = nRT; assuming P,T a standart conditions ( 1 atm, 298 K )

Therefore n/V = P/RT

Step 1: Calculate the moles of reactants and products

n = (1 atm x 2.00 L)/(0.0821 atm.L/K.mol x 298K)

n = 0.082 mol

mol of SO3 = 0.082 mol = mol of SO2 = mol of O2/2

Step 2: Calculate the concentrations of reactants and products

[SO3] = 0.082 mol/2.00L = 0.041 M

[SO2] = 0.041 M

[O2] = (0.082 mol/2.00 L)/2 = 0.0205 M

Step 3: Calculate Q

Q = [(0.0205)(0.041)^2]/(0.041)^2

Q = 0.0205

Q < Kc

Answer

By decreasing the volume, the reaction will favour forward reaction, Q < Kc