Question
Question: What is minimum mass of \(CaC{O_3}\) below which it decomposes quickly required to \({K_c} = 0.05mol...
What is minimum mass of CaCO3 below which it decomposes quickly required to Kc=0.05mol/litre establish equilibrium in a 6.5 litre container for the reaction:
CaCO3⇌CaO+CO2
a) 32.5g
b) 24.6g
c) 40.9g
d) 8.0g
Solution
CaCO3 is the chemical formula for Calcium carbonate which is a carbonic salt of calcium.Chemical reactions are said to either proceed in the forward direction or in the backward direction.However when the rate of forward reaction is same as the rate of backward reaction the reaction is said to have attained equilibrium.
Complete answer:
The most important aspect of a reaction in equilibrium is that the concentrations of the reactants and products remain the same throughout.But in order to set a reaction in equilibrium the concentrations of the reactants needs to be adjusted.This adjustments are subjects to parameters like pressure,volume of the container etc.
In the above chemical reaction calcium carbonate is undergoing decomposition reaction to give out Calcium oxide and Carbon dioxide gas.Out of all the species involved only carbon dioxide is in gaseous phase and the rest are in solid phase.
Here the equilibrium constant Kc is given as 0.05mol/litre
But,
Kc=[CO2]
We know that,
Volume ofCO2Moles ofCO2=0.05
Hence the number of moles of CO2 required is 0.05×6.5=0.325
Therefore the number of moles present should be slightly greater than 0.325
The minimum weight of calcium carbonate is then given by 0.325×100=32.5g
Hence the correct option is a)32.5g.
Note:
A chemical reaction in which a reactant breaks down into two or more different types of simpler products is known as decomposition reaction.Calcium carbonate is most commonly found in rocks as the minerals calcite and aragonite and is the main component of eggshells, snail shells, seashells and pearls.