Question

For the following reaction, $1g$ mole of $CaC{O_3}$ ​ is enclosed in $5L$ container.
$CaC{O_3}\left( s \right)\, \to \,CaO\left( s \right)\, + \,C{O_2}\left( g \right)$
$Kp\, = \,1.16$ at $1073\,K$, then percentage dissociation of $CaC{O_3}$​ is:
$A.\,0\%$
$B.\,6.58\%$
$C.\,65\%$
$D.\,100\%$

Answer 1

Calcium carbonate is a chemical compound with the formula $CaC{O_3}$ . It is a common substance found in rocks in the form of the minerals calcite and aragonite (especially in the form of limestone, which is a type of sedimentary rock made up mostly of calcite) and is the main component of eggs, snail shells, seashells and pearls. Calcium carbonate is the active ingredient in agricultural lime and is created when calcium ions in hard water react with carbonate ions to create limestone. It has medical use as a calcium supplement or as an antacid, but excessive consumption can be dangerous and lead to poor digestion.

Complete step-by-step answer: In this question we need to find the dissociation per cent of calcium carbonate.
So, first, let’s see the given values in the question;
The initial moles of calcium carbonate CaC{O_3}$$$$ = \,1\,g/mol
Volume of container $= \,5\,L$
Equilibrium constant $Kp\, = \,1.16$
Temperature $T\, = \,1073\,K$
$CaC{O_3}\left( s \right)\, \to \,CaO\left( s \right)\, + \,C{O_2}\left( g \right)$
Initial moles will be $1\,g/mol$ for $CaC{O_3}$
Initial moles will be $0$ for $CaO$
Initial moles will be $0$ for $C{O_2}$
The moles at equilibrium at $t$ time for $CaC{O_3}$ will be $1 - x$
The moles at equilibrium at $t$ time for $CaO$will be $x$
The moles at equilibrium at $t$ time for $C{O_2}$will be $x$
At equilibrium the concentration for $CaC{O_3}$ will be $\dfrac{{1 - x}}{5}$
At equilibrium the concentration for $CaO$ will be $\dfrac{x}{5}$
At equilibrium the concentration for $C{O_2}$ will be $\dfrac{x}{5}$
As we know that
${\text{Kc}}\,{\text{ = }}\,\dfrac{{{\text{[product]}}}}{{{\text{[reactant]}}}}$
$Kc\, = \,\dfrac{x}{5}$
$Kp\, = \,Kc{(RT)^{\Delta n}}$
Let’s substitute the entire values in the above equation;
We get,
$1.16\, = \,\dfrac{x}{5}{(0.0821\, \times \,1073)^1}$
$x\, = \,\dfrac{{1.16\, \times \,5}}{{0.0821\, \times \,1073}}$
$= \,0.0658$
Now, we need to convert in percentage therefore, we need to multiply with hundred;
$= \,0.0658\, \times \,100$
So, the obtained answer;
$= \,6.58\,\%$
The percentage of dissociation of calcium carbonate is $\,6.58\,\%$

Therefore, the correct option is $B.\,6.58\%$

Note: $Kp\,$ is the equilibrium constant which is calculated from the partial pressures of a chemical reaction. This is used to express the relationship between product pressures and reactant pressures. It is a unitless number, although it relates to the pressures.