Q17.97CP - Chemistry: Molecular Nature Of Matter And Change

Question

A gaseous mixture of $\mathbf{10}\mathbf{.}\mathbf{0}$ volumes of   ${\mathbf{\text{CO}}}_{\mathbf{2}}\mathbf{,}\mathbf{1.00}$  volume of unreacted data-custom-editor="chemistry" ${\mathbf{O}}_{\mathbf{2}}$ , and $\mathbf{50}\mathbf{.}\mathbf{0}$ volumes of unreacted ${\mathbf{N}}_{\mathbf{2}}$  leaves an engine at $\mathbf{4}\mathbf{.0}\mathbf{ }\mathbf{\text{atm}}$ and 800K. Assuming that the mixture reaches equilibrium, what are

(a) the partial pressure and

(b) The concentration (in picograms per litre,$\mathbf{\text{pg}}\mathbf{/}\mathbf{\text{L}}$  ) of  data-custom-editor="chemistry" $\mathbf{\text{CO}}$ in this exhaust gas? data-custom-editor="chemistry" $\mathbf{2}{\mathbf{\text{CO}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{\text{CO}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}{\mathbf{K}}_{\mathbf{\text{p}}}\mathbf{=}\mathbf{1}\mathbf{.4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{28}}$ at  data-custom-editor="chemistry" $\mathbf{800}\mathbf{.}\mathbf{\text{\hspace{0.17em}K}}$ (The actual concentration of  data-custom-editor="chemistry" $\mathbf{CO}$ in exhaust gas is much higher because the gases do not reach equilibrium in the short transit time through the engine and exhaust system.)

Step-by-Step Solution

Answer

(a) The partial pressure of  $\text{CO}$  at equilibrium is $3\cdot {10}^{14} \text{atm}$ .

(b) The concentration of $\mathrm{CO}$  is $0.013\text{pg}/\text{L}$ .

 

Step 1: Definition of equilibrium

Chemical equilibrium is the state in which no net change in the amounts of reactants and products occurs during a reversible chemical reaction.

Step 2: Find the partial pressure

a)

Given reaction:

 $2{\text{CO}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\text{CO}\left(\mathrm{g}\right)+{\text{O}}_2\left(\mathrm{g}\right)$

The values are:

 ${\mathrm{K}}_{\text{p}}=1.4\cdot {10}^{-28}\text{ at }800 \text{K}$

 ${\text{V}}_{{\text{CO}}_2}=10 \text{L}$

 $V_{{\text{O}}_2}=1 \text{L}$

 ${\text{V}}_{{\text{N}}_2}=50 \text{L}$

Firstly we will calculate the total volume of the gases by knowing that the amount of  $\mathrm{CO}$ is very small SO:

$\text{Total volume }={\text{V}}_{{\text{CO}}_2}+{\text{V}}_{{\text{O}}_2}+{\text{V}}_{{\text{N}}_2}$

 $\text{Total volume }=10 \text{L}+1 \text{L}+50 \text{L}$

 $\text{Total volume }=61 \text{L}$

Now we will calculate the equilibrium partial pressure of  ${\text{CO}}_2,{\text{O}}_2$and ${\mathrm{N}}_2$  :

${\text{P}}_{{\text{CO}}_2}=\frac{\text{ Volume }}{\text{ Total volume }}\cdot \text{ Total pressure }$

${\text{P}}_{{\text{CO}}_2}=\frac{10}{61}\cdot 4 \text{atm}$

${\text{P}}_{{\text{CO}}_2}=0.6557 \text{atm}$

${\text{P}}_{{\text{O}}_2}=\frac{\text{ Volume }}{\text{ Total volume }}\cdot \text{ Total pressure }$

${\text{P}}_{{\text{O}}_2}=\frac{1}{61}\cdot 4 \text{atm}$

${\text{P}}_{{\text{O}}_2}=0.06557 \text{atm}$

${\text{P}}_{{\text{N}}_2}=\frac{\text{ Volume }}{\text{ Total volume }}\cdot \text{ Total pressure }$

${\text{P}}_{{\text{N}}_2}=\frac{50}{61}\cdot 4 \text{atm}$

${\text{P}}_{{\text{N}}_2}=3.279 \text{atm}$

Now we will use the reaction for the equilibrium constant to express and calculate the partial pressure of CO at equilibrium:

${\mathrm{K}}_{\text{p}}=\frac{{\left({\mathrm{P}}_{\text{CO}}\right)}^2{\mathrm{P}}_{{\text{O}}_2}}{{\left({\mathrm{P}}_{{\text{CO}}_2}\right)}^2}$

$1.4\cdot {10}^{28}=\frac{{\left(P_{\text{CO}}\right)}^2\cdot 0.06557 \text{atm}}{{(0.6557 \text{atm})}^2}$

${\left({\mathrm{P}}_{\text{CO}}\right)}^2=\frac{1.4\cdot {10}^{28}\cdot {(0.6557 \text{atm})}^2}{0.06557 \text{atm}}$

${\left({\mathrm{P}}_{\text{CO}}\right)}^2=9.1798\cdot {10}^{28}$

${\mathrm{P}}_{\text{CO}}=\sqrt{9.1798\cdot {10}^{28}}$

${\mathrm{P}}_{\text{CO}}=3\cdot {10}^{14} \text{atm}$

Therefore, the required value is  ${\mathrm{P}}_{\text{CO}}=3\cdot {10}^{14} \text{atm}$.

Step 3: Calculate the concentration of CO

Let us solve the given problem.

${\left[\text{CO}\right]}_{\text{equilibrium }}=?$ - The concentration of  $\mathrm{CO}$ at equilibrium to moles per litre

It is expressed as:

 data-custom-editor="chemistry" ${\left[\text{CO}\right]}_{\text{eq }}=\frac{\mathrm{n}}{\mathrm{V}}$

 

 ${\left[\text{CO}\right]}_{\text{eq }}=\frac{\mathrm{P}}{\text{RT}}$

 ${\left[\text{CO}\right]}_{\text{eq }}=\frac{3\cdot {10}^{14} \text{atm}}{0.0821 \text{L}\cdot \text{atm}/\text{mol}\cdot \text{K}\cdot 800 \text{K}}$

${\left[\text{CO}\right]}_{\text{eq }}=4.61\cdot {10}^{16} \text{mol}/\text{L}\cdot \frac{28.01\text{gCO}}{1\text{molCO}}\cdot \frac{1\text{pg}}{{10}^{12} \text{g}}$

${\left[\text{CO}\right]}_{\text{eq }}=0.013\text{pg}/\text{L}$ .

Therefore, the required value is ${\left[\text{CO}\right]}_{\text{eq }}=0.013\text{pg}/\text{L}$ .