Question

For the reaction $CaC{O_3}\left( s \right) \rightleftharpoons CaO\left( s \right)$ $Kp = 1.16atm$ at ${800^ \circ }C$ . If $20g$ of $CaC{O_3}$ was put into a $10L$ container and heated at ${800^ \circ }C$ . The percentage of the $CaC{O_3}$ that would remain unreacted at equilibrium $\left( {R = 0.082Latm/mol/K} \right)$ is about.

Answer 1

Hint:We know that the ideal gas law is the condition of a speculative ideal gas. It is a decent estimation of the conduct of numerous gases under numerous conditions, despite the fact that it has a few impediments
The ideal gas equation is,
$PV = nRT$
Where P is the pressure in the atmosphere.
V is the volume of gas in a liter.
n is the number of moles.
R is a universal gas constant.
T is the temperature.

Complete step by step answer:
We know that $Kp = PC{O_2}$
It is given that value of $Kp = 1.16atm$
The temperature is ${800^ \circ }C$ .
The volume is $10L$ .
The gas constant is $R = 0.082Latm/mol/K$
Now we can calculate the number of moles reacted using the ideal gas equation,
\[{\text{n = }}\dfrac{{PV}}{{RT}}\]
\[\dfrac{{\text{W}}}{M}{\text{ = }}\dfrac{{1.16 \times 10}}{{0.082 \times 1073}}\]
\[{\text{W = }}\dfrac{{1.16 \times 10 \times 44}}{{0.082 \times 1073}}\]
On simplifying we get,
\[{\text{W = 5}}{\text{.8g}}\]
The balanced equation is,
$CaC{O_3}\left( s \right) \rightleftharpoons CaO\left( s \right) + C{O_2}$
The mass of calcium carbonate reacted can be calculated as,
$\dfrac{x}{{100}} = \dfrac{{5.8g}}{{44}}$
On simplifying we get,
$x = 13.1g$
The mass of calcium carbonate unreacted $ = 20 - 13.1 = 6.9$
The percentage of calcium carbonate unreacted $ = \dfrac{{6.9}}{{20}} \times 100 = 34.5\% $

Additional information:
If the gas obeys an ideal gas equation then the pressure is given by,
${\text{P = }}\dfrac{{{\text{nRT}}}}{{\text{V}}} \to (1)$
If the volume is doubled and the temperature is halved then the equation becomes,
${\text{P = }}\dfrac{{{\text{nRT/2}}}}{{{\text{2V}}}}$
${\text{P = }}\dfrac{{{\text{nRT}}}}{{{\text{4V}}}} \to 2$
From equation 1 ${\text{P = }}\dfrac{{{\text{nRT}}}}{{\text{V}}}$ then the equation 2 becomes,
${\text{P = }}\dfrac{{\text{P}}}{{\text{4}}}$
Thus, if the volume is doubled and the temperature is halved then the pressure of the system decreases by four times.

Note:
We know that,
$Density = \dfrac{{mass}}{{volume}}$
Assuming mass is equal to the number of moles in ideal gas.
$Density = \dfrac{n}{{Volume}}$
The ideal gas equation is,
${\text{PV = nRT}}$
The number of moles can be calculated as,
${\text{n = }}\dfrac{{{\text{PV}}}}{{{\text{RT}}}}$
Substituting the value of n in density equation,
\[{\text{Density = }}\dfrac{{PV}}{{RTV}}\]
\[{\text{Density = }}\dfrac{{\text{P}}}{{{\text{RT}}}}\]
\[{\text{Density}} \propto \dfrac{{\text{1}}}{{\text{T}}}\]
It is clear that density is inversely proportional to temperature. Thus, as the density of the gas decreases temperature increases.