Question

5. The following decomposition of calcium carbonate has Kp -1.16 at 800.°C CaCOs(s)CaO(s) + CO() Suppose a 100-g sample of solid CaCOs is placed in a sealed, evacuated 5 0-L flask, and the sample decomposes to reach equilibrium at 800 C according to the above equation. What percentage (by mass) of the solid CaCO, have decomposed at equilibrium? (a) 66% (b) 34% (c) 13% (d) 87%

Answer 1

Given that,

K_p = 1.16

and K_p = K_c (RT)^$\Delta$n

$\Delta$n = moles of gaseous product - moles of gaseous reactant = 1-0 = 1

Temperature (T) = 800^oC = 800 + 273 = 1073 K

1.16 = K_c ( 0.08206 L.atm/mol.K * 1073K) = 88.1K_c

K_c = 0.0132

Mass of CaCO₃ = 10 g

Molar mass of CaCO₃ = 100 g/mol

Moles of CaCO₃ = Mass/ molar mass = 10/100 = 0.1 mol

Reaction is:

	CaCO3(s)	$\rightleftharpoons$	CaO(s) +	CO2(g)
Initial (mol)	0.1		0	0
Change(mol)	-x		+x	+x
equilibrium (mol)	0.1-x		x	x

K_c = [CO₂] ....(activities of solids is 1. So not included in K_c)

0.0132 = x

Moles of CaCO₃ decomposed = 0.1 - 0.0132 = 0.0868 mol

Mass of CaCO₃ = Moles * molar mass = 100 g/mol * 0.0868 mol = 8.68 g

Thus, percentage of CaCO₃ decomposed = 8.68/10*100% = 87%

option D