Question

Question: At 900.^oC, K_p = 1.04 for the reaction

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

At a low temperature dry ice (solid CO₂), calcium oxide, and calcium carbonate are introduced into a 50.0-L reaction chamber. The temperature is raised to 900.^oC. For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium?

a. 655 g of CaCO₃, 95.0 g of CaO, 58.4 g of CO₂

b. 780 g of CaCO₃, 1.00 g of CaO, 23.76 g of CO₂

c. 0.14 g of CaCO₃, 5000 g of CaO, 23.76 g of CO₂

d. 715 g of CaCO₃, 813 g of CaO, 4.82 g of CO₂

Short answer

a. The initial amount of calcium oxide will increase.

b. The initial amount of calcium oxide will remain the same.

c. The initial amount of calcium oxide will remain the same.

d. The initial amount of calcium oxide will decrease.

Step-by-step solution

Find the initial amount of calcium oxide change in part a.

The expression for the reaction quotient, Q, for the decomposition reaction of calcium carbonate is written below:

$\mathrm{Q}={\mathrm{P}}_{{\mathrm{CO}}_{\mathrm{2}}}$ ...(1)

Use the below ideal gas law to calculate the partial pressure of carbon dioxide.

${\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{{\text{n}}_{{\text{CO}}_{\text{2}}}\text{RT}}{\text{V}}$ ...(2)

Here,

${\text{P}}_{{\text{CO}}_{\text{2}}}$is the partial pressure of carbon dioxide.

${\text{n}}_{{\text{CO}}_{\text{2}}}$is the number of moles of carbon dioxide.

V is the volume.

T is the temperature.

R is the gas constant 0.08206Latm/molK.

Insert Equation (2) into Equation (1).

$\text{Q =}\frac{{\text{n}}_{{\text{CO}}_{\text{2}}}\text{RT}}{\text{V}}$

Since ${\text{n}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{{\text{m}}_{{\text{CO}}_{\text{2}}}}{\text{M}}$

$\mathrm{Q}=\frac{{\displaystyle \frac{{\mathrm{m}}_{{\mathrm{co}}_2}}{\mathrm{M}}}\times \mathrm{RT}}{\mathrm{V}}$ ...(3)

Substitute the necessary values into Equation (3).

$$\begin{gathered} \mathrm{Q}=\frac{{\displaystyle \frac{58.4\mathrm{g}}{44.01\mathrm{g}/\mathrm{mol}}}\times 0.08206\mathrm{Latm}/\mathrm{molK}\times 1173\mathrm{K}}{50.0\mathrm{L}} \\ =2.55 \end{gathered}$$

The equilibrium constant, K_p value, is given as 1.04.

The Q value is greater than the K_p value.So, the system will shift to the left. Hence, the initial amount of calcium oxide will increase.

Find the initial amount of calcium oxide change in part b.

The mixture in part b is given as 780 g of CaCO₃, 1.00 g of CaO, and 23.76 g of CO₂.

Substitute these values into Equation (3).

$$\begin{gathered} \text{Q=}\frac{{\displaystyle \frac{23.76\mathrm{g}}{44.01\mathrm{g}/\mathrm{mol}}\times 0.08206\mathrm{Latm}/\mathrm{molK}\times 1173\mathrm{K}}}{50\mathrm{L}} \\ =1.04 \end{gathered}$$

The mass of carbon dioxide in part c is the same as in part b.Hence, the initial amount of calcium oxide will remain the same.

 Step 4: Find the initial amount of calcium oxide change in part d.

The mixture in part d is given as 715 g of CaCO₃, 813 g of CaO, and 4.82 g of CO₂.

Substitute these values into Equation (3).

$$\begin{gathered} \mathrm{Q}=\frac{{\displaystyle \frac{4.82\mathrm{g}}{44.01\mathrm{g}/\mathrm{mol}}}\times 0.08206\mathrm{Latm}/\mathrm{molK}\times 1173\mathrm{K}}{50.0\mathrm{L}} \\ =0.211 \end{gathered}$$

The calculated Q value is less than the K_p value. So, the system will shift to the right side of the reaction. Hence, the initial amount of calcium oxide will decrease.