Question

For the reaction below, K_p = 1.16 at 800.^oC.

${\mathbf{CaCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\stackrel{}{\mathbf{\rightleftharpoons }}\mathbf{CaO}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

If a 20.0-g sample of ${\mathbf{CaCO}}_3$ is put into a 10.0-L container and heated to $\mathbf{800}{.}^o\mathbf{C}$ , what percentage by mass of the ${\mathbf{CaCO}}_3$ will react to reach equilibrium?

Short answer

The mass percent of calcium carbonate reacted to reach the equilibrium is 66.0%.

Step-by-step solution

The expression for equilibrium constant

The equation for the equilibrium constant, K_p of the reaction in terms of partial pressures is written below:

${\text{K}}_{\text{P}}{\text{=P}}_{{\text{CO}}_{\text{2}}}$ …(1)

Since the activities of pure liquids and solids are taken as unity.

The equilibrium constant value of the reaction is given as, ${\text{K}}_{\text{P}}\text{=1.16}$.

According to Equation (1), the partial pressure of carbon dioxide at equilibrium is 1.16 atm.

The number of moles of  CO2

The number of moles of carbon dioxide gas at equilibrium can be obtained by using the ideal gas equation.

$\text{n=}\frac{\text{PV}}{\text{RT}}$

Here,

n is the number of moles.

P is the pressure.

V is the volume.

T is the temperature.

R is the gas constant, i.e., $\text{0.08206\hspace{0.17em}L\hspace{0.17em}atm/mol\hspace{0.17em}K}$.

$\begin{array}{c}\text{n=}\frac{\text{1.16\hspace{0.17em}atm×10.0\hspace{0.17em}L}}{\text{0.08206\hspace{0.17em}L\hspace{0.17em}atm/mol\hspace{0.17em}K×1073\hspace{0.17em}K}}\\ \text{=0.132\hspace{0.17em}mol}\end{array}$

Hence, the number of moles of carbon dioxide gas at equilibrium is obtained to be $\text{0.132\hspace{0.17em}mol}$.

The mass of CaCO3

Use the below expression to calculate the mass of calcium carbonate reacted.

${\text{0.132molCO}}_{\text{2}}\text{×}\frac{{\text{1molCaCO}}_{\text{3}}}{{\text{1molCO}}_{\text{2}}}\text{×}\frac{{\text{100.09\hspace{0.17em}g\hspace{0.17em}CaCO}}_{\text{3}}}{{\text{1molCaCO}}_{\text{3}}}{\text{=13.2\hspace{0.17em}g\hspace{0.17em}CaCO}}_{\text{3}}$

Hence, the mass of calcium carbonate reacted is 13.2 g.

The mass% of  that reacted to reach the equilibrium

The following equation is used to calculate the mass percent of calcium carbonate reacted to reach the equilibrium.

$\begin{array}{c}{\text{Mass\hspace{0.17em}percent\hspace{0.17em}of\hspace{0.17em}CaCO}}_{\text{3}}\text{\hspace{0.17em}reacted=}\frac{{\text{Mass\hspace{0.17em}of\hspace{0.17em}CaCO}}_{\text{3}}\text{\hspace{0.17em}reacted}}{{\text{Initial\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}CaCO}}_{\text{3}}}\text{×100\%}\\ \text{=}\frac{\text{13.2\hspace{0.17em}g}}{\text{20.0\hspace{0.17em}g}}\text{×100\%}\\ \text{=66.0\%}\end{array}$

Therefore, the mass percent of calcium carbonate reacted to reach the equilibrium is 66.0%.