Question

When 20g of CaCo3 were put into 10 litre flask and heated to 800°C,30% of CaCo3 remained unreacted at equilibrium K_p for decomposition of CaCo3 will be -------
1)1.145atm 2)1.231 atm
2)2.146 atm 4)3.145 atm

Answer 1

The decomposition of calcium carbonate will follow

CaCO3(s) --> CaO(s)+CO2(g)

For this reaction Kp= P_Co2 because others are solids.

So we need to find the pressure of carbon dioxide

It is given that 30% of 20g CaCO3 remains unreacted so 70% must be reacted

70% of 20g = 14g

Number of Moles of CaCO3= 14g/100gmol-1(Molar mass of CaCO3=100g)

= 0.14moles

In according to the equation we will get 0.14 mole of CO2

Pressure of CO2 can be found using ideal gas equation

PV=nRT

Px10 litre = 0.14moles x 0.082 x (800+273)

Which will give pressure as 0.14x0.082x1073/10 = 1.23 atm

This is the Kp of equilibrium.