Question

For the reaction below, Kp 5 1.16 at 800.8C. CaCO3(s) 34 CaO(s) 1 CO2(g) If a 20.0-g sample of CaCO3 is put into a 10.0-L container and heated to 800.8C, what percentage by mass of the CaCO3 will react to reach equilibrium?

Answer 1

Answer:

The mass percentage of calcium carbonated reacted is 2.5%.

Explanation:

The reaction is:

$CaCO_{3}(s)--->CaO(s)+CO_{2}(g)$

Thus the Kp of the equilibrium will be:

Kp = partial pressure of carbon dioxide [as the other are solid]

Moles of calcium carbonate initially present = $\frac{mass}{molarmass}=\frac{20}{100}=0.2$

Let us apply ICE table to the equilibrium given:

                        $CaCO_{3}(s)--->CaO(s)+CO_{2}(g)$

Initial                       0.2                       0          0

Change                 -x                            +x        +x

Equilibrium           0.2-x                         x          x

Kp = partial pressure of carbon dioxide

Kp = Kc(RT)ⁿ

where n = difference in the number of moles of gaseous products and reactants

for given reaction n = 1

R = gas constant = 8.314 J /mol K

T = temperature = 800 ⁰C = 1073 K

Putting values

Kc =$\frac{Kp}{RT}=\frac{1.16}{8.314X1073}=1.3X10^{-4}$

Kc = $\frac{[CO_{2}][CaO]}{[CaCO_{3}]}= \frac{x^{2} }{(0.2-x)}=1.3X10^{-4}$

$1.3X10^{-4}(0.2-x)=x^{2}$

$x^{2} = 0.26X10^{-4}-1.3X10^{-4}x$

On calculating

x =  0.005

where x = the moles of calcium carbonate dissociated or reacted.

Percentage of the moles or mass reacted = $\frac{molesreacted X100}{initialmoles}=\frac{0.005X100}{0.2}=2.5$%