Answers
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.
.