Question

Carbon monoxide gas may be formed as follows. C0_2(g) + C(s) 2 CO(g) K_p = 5.78 at 1200 K 3.46 g of C0_2,1.98 g of C(s) and 7.22 g of CO(g) are placed in a 12.2 L container and heated to 1200 K. Calculate the total pressure (in bar) at equilibrium. All mathematical steps must be shown. If the volume of the vessel is allowed to expand to 40.0 L when the system is already at equilibrium, would you predict that the moles of CO(g) would increase, decrease or stay the same? Explain. If the system is at equilibrium and the temperature is decreased to 800 K, would you predict that the moles of CO(g) would increase, decrease or stay the same? Explain.

Answer 1

a.) We will only consider the gases

Weight of CO2 = 3.46 g

moles of CO2 = 3.46/44 = 0.07863 moles

Weight of CO = 7.22 g

moles of CO = 7.22/28 = 0.25785 moles

mole fractions,

X1(CO2) = 0.07863/( 0.07863+0.25785) = 0.2336

X2(CO) =  0.25785/( 0.07863+0.25785) = 0.7664

So,

Let total pressure be Pt

So, partial pressures are :

P1(CO2) = X1*Pt = 0.2336 Pt

P2(CO) = X2*Pt = 0.7664Pt

Now,

For reaction :

CO2 + C --- > 2 CO

Kp = [P2(CO)]^2 / [P1(CO2)]

5.72 = (0.7664Pt)^2/ (0.2336 Pt )

Pt = 5.72 * 0.2336 / (0.7664)^2 = 2.2748 atm

Hence, total pressure at equilibrium is 2.2748 atm

b.)

  By Le-chatliers principle ,

If the volume of the system is incresead, the equilibrium should shift such that more no of molecules of gases are formed to occupy the incresed space. So, the equlibrium should shift to the side where the no of moles of gases are more in total.

Here, the no of moles of gases is 2 in product side whereas in reactant side its 1.

So, when the volume is increased to 40 L, the equilibrium shifts towards right.

and more no of moles of CO are formed i.e. moles of CO will increase.

c.)

By Le-chatliers principle ,

If the tmperaure of the system isdecreased, the equilibrium should shift such that more no of molecules of gases decreases so that it occupies less space. as PV = nRT, as temperature is directly proportional to Volume. it increase decrease has same effect.So, the equlibrium should shift to the side where the no of moles of gases are less in total.

Here, temperature is dcreased from 1200 K to 800 K

Here, the no of moles of gases is 2 in product side whereas in reactant side its 1.

So, the equilibrium shifts towards left.

So, the no of moles of CO decreases.