Question

A 50 g $CaC{{O}_{3}}$are heated to 1073 K in a 5L vessel. What percentage of $CaC{{O}_{3}}$would decompose at equilibrium? The value of ${{K}_{p}}$for the reaction. $CaC{{O}_{3}}(s)\rightleftharpoons CaO(s)+C{{O}_{2}}(g)$is 1.15 atm at 1073 K.

Answer 1

Hint:The equilibrium constant of a chemical reaction is represented by the symbol K which gives us the relationship between products and reactants when a chemical reaction reaches equilibrium. It can also be defined as the ratio between the amounts of reactant to the amount of product.

Complete step-by-step answer:Equilibrium is that state of reaction where the rate of forward reaction is equal to the rate of backward reaction. Here the term ${{K}_{p}}$indicates the equilibrium constants formula in terms of partial pressure.
According to the question
Amount of $CaC{{O}_{3}}$= 50 g
Molar mass of $CaC{{O}_{3}}$= $40+12\times 3\times 16$= 100 g
Number of moles of $CaC{{O}_{3}}$= $\frac{0.065}{5}\times 100=1.3%$$\frac{Given\ \text{mass}}{molar\ \text{mass}}$
$\frac{50}{100}=0.5\ \text{mole}$
$CaC{{O}_{3}}(s)\rightleftharpoons CaO(s)+C{{O}_{2}}(g)$
${{K}_{p}}$= ${{P}_{C{{O}_{2}}}}=1.15\ \text{atm}$
Volume = 5 L, Temperature = 1073 K
Ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behavior of many gases under many conditions and the equation of ideal gas law can be written as:
$PV=nRT$
Where P is pressure, v is volume, R is gas constant, T is temperature in kelvin and n is amount of ideal gas measured in terms of moles.
$n=\frac{{{P}_{C{{O}_{2}}}}V}{RT}$
$n=\frac{1.15\times 5}{0.821\times 1073}=0.065$
% of $CaC{{O}_{3}}$decompose = $\frac{0.065}{5}\times 100=1.3%$
Thus we can say that 1.3% of $CaC{{O}_{3}}$ would decompose at equilibrium.

Note:Equilibrium constant can be affected by various factors like change in concentration of any reactant or product, change in temperature and pressure of the system. Or by adding inert gas or catalyst the value of equilibrium constant also changed.