100gm of CaCO3 is heated in a vessel of 100 l capacity at 1000k. CaCO3 decomposes to form CaO and CO2 ..if kp for the equilibrium CaCO3 <----> CaO + CO2 is 7.5 atm, then find the % of CaCO3 unreacted
1CaCO3 -> CaO+CO2
P-x x x
x2/P-x=7.5 -- - -- (i)
PV=nrT
n=1(100/100)
R=.08
T=1000
V=100
hence P=1*.08*1000/100=0.8
we can calculate x from (i)
hence we get P-x
percentage of CaCo3 unreacted= n2/1*100
n2*RT=(P-x)V
putting the same values we get the answer
1since CO2 is the only gas in this equilibrium, Kp = pressure of CO2
P=7.5, V=100L, R=0.0821, T=1000K, find no. of moles of CO2.
find initial no. of moles of CaCO3 = n1
no. of CO2 moles = no. of CaCO3 moles reacted = n2
no. of CaCO3 moles left is n1-n2.
%unreacted = [(n1 - n2)/n1] * 100
1yes...got it !!... guys !!... thanks !!.. caco3 reacts 100% here....
hence unreacted = 0% :P