Question

\[CaC{{O}_{3(s)}}\rightleftharpoons Ca{{O}_{(s)}}+C{{O}_{2(g)}};{{K}_{p}}=1.64 atm\] at 1000K
50g of \[CaC{{O}_{3(s)}}\]in a 10 litre closed vessel is heated to 1000K. Percentage of \[CaC{{O}_{3(s)}}\] that remains unreacted at equilibrium is:
\[(Given:R=0.082Latm{{K}^{-1}}mo{{l}^{-1}})\]
(a) 50
(b) 20
(c) 40
(d) 60

Answer 1

Hint: According to ideal gas law, at constant pressure and temperature, the volume of a particular is directly proportional to the number of moles of the gas. It explains the behaviour of gases, although it has its limitations.

Complete step by step solution:
According to the question, the reaction is given as \[CaC{{O}_{3(s)}}\rightleftharpoons Ca{{O}_{(s)}}+C{{O}_{2(g)}}\]

\[{{K}_{p}}\]=\[{{p}_{C{{O}_{2}}}}\] and \[{{K}_{c}}\]=\[[C{{O}_{2}}]\], given mass of \[CaC{{O}_{3}}\]=50g and Molecular mass= 100g/mol.
So we can calculate the number of moles of \[CaC{{O}_{3}}\]=\[\frac{Mass}{Mol.mass}\]=\[\dfrac{50}{100}\]=0.5
Next step is to apply ideal gas law
\[PV=nRT\]-(i) ,here P is the pressure of the gas, V is the volume given and R is Rydberg constant , n is the number of moles of calcium carbonate and T is the temperature.
Their values are given as: P=1.64atm, V=10L ,T= 1000k and \[R=0.082Latm{{K}^{-1}}mo{{l}^{-1}}\]
From solving the equation (i) we get n (number of moles of \[CaC{{O}_{3}}\]reacted) by substituting the above values. So, we get
\[1.64\times 10=n\times 0.082\times 1000\]
\[n=0.2\]
Therefore, the number of moles of unreacted \[CaC{{O}_{3}}\]= 0.5-0.2 = 0.3.
The percentage of unreacted \[CaC{{O}_{3}}\] \[=\dfrac{0.3}{0.5}\times 100=60%\]

So, the correct answer is “Option D”.

Additional Information:
There are 3 fundamental gas laws which give the relation between the temperature, volume and pressure of gases. Boyle’s Law states as the pressure increases, volume of the gas decreases. Charles’s Law states that volume and temperature of gas is directly proportional and Avogadro Law states that the number of moles of gas is directly proportional to its volume. The combination of all these laws is the ideal gas law (\[PV=nRT\]). But ideal gas is just an assumption. The gases which show deviation from this law are called real gases. All the gases are real gases. Therefore, modification has been done to ideal gas law which is applicable for real gases by introducing a Z also known as compressibility factor. Thus, the gas law becomes \[PV=ZnRT\]

Note: When n is calculated from the gas equation, we get the number of moles of \[CaC{{O}_{3}}\] which has reacted. To get the number of moles of unreacted \[CaC{{O}_{3}}\] , we must subtract it from the total number of moles of \[CaC{{O}_{3}}\].