Q17.91CP - Chemistry: Molecular Nature Of Matter And Change

Question

When $\mathbf{\text{0.100 mol}}$ of ${\mathbf{\text{CaCO}}}_{\mathbf{\text{3}}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$ and $\mathbf{\text{0.100 mol}}$ mol of $\mathbf{\text{CaO(s)}}$  are placed in an evacuated sealed $\mathbf{\text{10.0-L}}$ container and heated to $\mathbf{\text{385 K}}$ , ${\mathbf{\text{P}}}_{{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}}\mathbf{\text{=0.220 atm}}$ after equilibrium is established:

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

An additional $\mathbf{\text{0.300 atm}}$ of  ${\mathbf{\text{CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$is pumped in. What is the total mass (in  $\mathbf{g}$ ) of ${\mathbf{\text{CaCO}}}_{\mathbf{\text{3}}}$ after equilibrium is re-established?

Step-by-Step Solution

Answer

The total mass of ${\text{CaCO}}_{\text{3}}$ after equilibrium is re-established is ${\text{12.5 g CaCO}}_{\text{3}}$ .

Step 1: Concept Introduction

Chemical equilibrium is the state of a system in which the concentration of the reactant and the concentration of the products do not change over time and the system's attributes do not change.

Step 2: Calculation for number of moles

The reaction given is –

${\text{CaCO}}_3\text{(S)}\rightleftharpoons {\text{CaO(s)+CO}}_2\text{(g)}$

First write the expression for the equilibrium constant of the reaction in terms of partial pressures of the reactants. It should only include gaseous reaction components.

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}\frac{{\text{P}}_{\text{products}}}{{\text{P}}_{\text{reactants}}}\\ {\text{K}}_{\text{p}}{\text{=P}}_{{\text{CO}}_{\text{2}}}\end{array}$

Therefore, ${\mathrm{K}}_{\mathrm{P}}$  is just equal to the partial pressure  ${\text{CO}}_2$.

At the first equilibrium, the moles of ${\text{CO}}_2$  can be solved using the ideal gas law equation.

$\begin{array}{c}\mathrm{PV}=\mathrm{nRT}\\ \frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\\ \text{=}\frac{\text{(0.220 atm)(10.0 L)}}{\left(\text{0.0821}\frac{\text{ atm ×L}}{\text{mol×K}}\right)\text{(385K)}}\\ {\text{n=0.0696 mol CO}}_{\text{2}}\end{array}$

Using stoichiometry, we can solve for the amount of moles used to produce  ${\text{0.0696 mol CO}}_{\text{2}}$.

$\begin{array}{c}{\text{mol CaCO}}_{\text{3}}{\text{=0.0696 mol CO}}_{\text{2}}\text{×}\frac{{\text{1 mol CaCO}}_{\text{3}}}{{\text{1 mol CO}}_{\text{2}}}\\ {\text{=0.0696 mol CaCO}}_{\text{3}}\end{array}$

Subtract it from the initial mol of  ${\text{CaCO}}_{\text{3}}$ present.

$\begin{array}{c}{\text{mol CaCO}}_{\text{3}}{\text{ left =(0.100-0.0696) mol CaCO}}_{\text{3}}\\ {\text{ =0.0304 mol CaCO}}_{\text{3}}\end{array}$

Step 3: Calculation for Mass

The reaction table is –

	${\mathrm{CaCO}}_3\text{(s)}$	$\mathrm{CaO}\text{(s)}$	${\mathrm{CO}}_2\text{(g)}$
Initial	-	-	$0.520$
Change	-	-	$+\mathrm{x}$
Equilibrium	-	-	$0.520+\mathrm{x}$

 

Substitute the equilibrium formula from our reaction table to the expression for the equilibrium constant in terms of partial pressures.

$\begin{array}{c}{\text{K}}_{\text{p}}{\text{=P}}_{{\text{CO}}_{\text{2}}}\\ \text{0.220=0.520+x}\\ \text{x=-0.300 atm}\end{array}$

The partial pressure of ${\mathrm{CO}}_2$  decreased by  $\text{0.300 atm}$. Therefore, ${\text{CaCO}}_{\text{3}}$  also increased with the same amount in terms of moles (given that their mole ratio is $1:1$  from the previous calculation). Solve for the number of moles of  ${\text{CO}}_2$ using the ideal gas law equation.

$\begin{array}{c}\mathrm{PV}=\mathrm{nRT}\\ \frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\\ \text{=}\frac{\text{(0.300 atm)(10.0 L)}}{\left(\text{0.0821}\frac{\text{ atm ×L}}{\text{mol×K}}\right)\text{(385K)}}\\ {\text{n=0.0949 mol CO}}_{\text{2}}\end{array}$

Since the mole ratio of   ${\text{CO}}_2$ and ${\text{CaCO}}_{\text{3}}$  is  $1:1$, then mol  ${\text{CO}}_2{\text{=mol CaCO}}_{\text{3}}$. Add it from the previous mol of  ${\text{CaCO}}_{\text{3}}$ calculated.

data-custom-editor="chemistry" $\begin{array}{c}{\text{mol CaCO}}_{\text{3}}{\text{ left =(0.0304+0.0949) mol CaCO}}_{\text{3}}\\ {\text{ =0.0125 mol CaCO}}_{\text{3}}\end{array}$

Solve for the mass of  data-custom-editor="chemistry" ${\text{CaCO}}_{\text{3}}$ using dimensional analysis.

data-custom-editor="chemistry" $\begin{array}{c}{\text{mass CaCO}}_{\text{3}}{\text{ left =0.0125 mol CaCO}}_{\text{3}}\text{×}\frac{{\text{100.09 g CaCO}}_{\text{3}}}{{\text{1 mol CaCO}}_{\text{3}}}\\ {\text{=12.5 g CaCO}}_{\text{3}}\end{array}$

 

Therefore, the value of mass is obtained as ${\text{12.5 g CaCO}}_{\text{3}}$ .