Question

The dissociation of calcium carbonate has an equilibrium constant of Kp = 1.16 at 1073 K . CaCO3(s) ⇄ CaO(s) + CO2(g) If you place 35.9 g of CaCO3 in a 9.56-L container at 1073 K, what is the pressure of CO2 in the container?

Answer 1

Answer:

$Pressure of CO_2 in the container=1.6 atm$

Explanation:

First balance the chemical equation:

$CaCO_3(s)$ ⇄  $CaO(s) + CO_2(g)$

two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container

Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp

$K_p=[CO_2]$

$[CO_2]=p$

$K_p=p$

$p=K_p = 1.16atm$

$Pressure of CO_2 in the container=1.6 atm$