Q17.79CP - Chemistry: Molecular Nature Of Matter And Change

Question

A mixture of \(5.00\) volumes of ${\mathbf{N}}_{\mathbf{2}}$ and 1.00 volume of ${\mathbf{O}}_{\mathbf{2}}$ passes slowly through a heated furnace. Assuming it reaches equilibrium at $\mathbf{1850}\mathbf{K}$ and $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{atm}$, the reaction is

${\mathbf{N}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{NO}\left(g\right)\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$

(a) What is the partial pressure of NO? 

(b) What is the concentration in micrograms per litre $\left(\mathrm{\mu g}/L\right)$ of $\mathbf{NO}$ in the mixture?

Step-by-Step Solution

Answer

1. The value is ${\mathrm{P}}_{\mathrm{NO}}=2.2\times {10}^{-2}$atm
2. The value is data-custom-editor="chemistry" $\left[\mathrm{NO}\right]=4.46\times {10}^{3 }\mathrm{\mu g}/\mathrm{L}$

Step 1: Definition of Concept

Reactions: A balanced chemical reaction equation shows the mole relationships of reactants and products, while a chemical reaction equation gives the reactants and products. The amount of energy involved in the reaction is frequently stated. Reaction stoichiometry is the study of the quantitative aspects of chemical reactions.

Step 2: Find the partial pressure of NO

(a)

Considering the given information: 

${\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2\mathrm{NO}\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{p}}=1.47\times {10}^{-4}$

$\mathrm{T}=1850 \mathrm{K}$

Identify the given values first.

5.00 volumes of and 1.00 volumes of means that the reactant has a 5 : 1 -mole ratio. Solve for the mole fraction of each reactant.

$$\begin{gathered} {\mathrm{X}}_{{\mathrm{N}}_2}=\frac{5.00\mathrm{mol} {\mathrm{N}}_2}{\left(5.00+1.00\right)\mathrm{mol}} \\ {\mathrm{X}}_{{\mathrm{N}}_2}=0.833 \\ {\mathrm{X}}_{{\mathrm{O}}_2}=\frac{1.00\mathrm{mol} {\mathrm{O}}_2}{\left(5.00+1.00\right)\mathrm{mol}} \\ {\mathrm{X}}_{{\mathrm{O}}_2}=0.167 \end{gathered}$$

At equilibrium, the total pressure of the reaction is ${\mathrm{P}}_{\mathrm{total}}=5.00 \mathrm{atm}$ . Multiply the mole ratios of each reactant to find the partial pressures of each.

$\begin{array}{rcl}{\mathrm{P}}_{{\mathrm{N}}_2}& =& {{\mathrm{X}}_{\mathrm{N}}}_2\times {\mathrm{P}}_{\mathrm{total}}\\ & =& \left(0.833\right)\left(5.00 \mathrm{atm}\right)\\ {\mathrm{P}}_{{\mathrm{N}}_2}& =& 4.165 \mathrm{atm}\\ {\mathrm{P}}_{{\mathrm{O}}_2}& =& {{\mathrm{X}}_{\mathrm{O}}}_2\times {\mathrm{P}}_{\mathrm{total}}\\ & =& \left(0.167\right)\left(5.00 \mathrm{atm}\right)\\ {\mathrm{P}}_{{\mathrm{O}}_2}& =& 0.835 \mathrm{atm}\end{array}$

Write the reaction table, next:

In terms of partial pressures, write the expression for the reaction's equilibrium constant.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=\frac{\left[\mathrm{product}\right]}{\left[\mathrm{reactant}\right]} \\ {\mathrm{K}}_{\mathrm{p}}=\frac{{\left[\mathrm{NO}\right]}^2}{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]} \end{gathered}$$

Using the reaction table and the given ${\mathrm{K}}_{\mathrm{p}}$, solve for x. Because the ${\mathrm{K}}_{\mathrm{p}}$ is small, we can remove the x from the denominator.

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{p}}& =& \frac{{\left[\mathrm{NO}\right]}^2}{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}\\ & =& \frac{{\left(2\mathrm{x}\right)}^2}{\left(4.165-\mathrm{x}\right)\left(0.8350-\mathrm{x}\right)}\\ {\mathrm{K}}_{\mathrm{p}}& =& \frac{4{\mathrm{x}}^2}{3.477}\\ \mathrm{x}& =& \sqrt{\frac{{\mathrm{K}}_{\mathrm{p}}\left(3.477\right)}{4}}\\ & =& \sqrt{\frac{\left(1.47\times {10}^{-4}\right)\left(3.477\right)}{4}}\\ \mathrm{x}& =& 0.0113\\ & & \end{array}$

Calculate the partial NO reaction.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{NO}}=2\mathrm{x}=2\left(0.0113\right) \\ {\mathrm{P}}_{\mathrm{NO}}=2.2\times {10}^{-2} \mathrm{atm} \end{gathered}$$

Step 3: Find the concentration in micrograms per litre

(b)

Considering the given information: 

We can use the relationship of pressure and moles, which is the ideal gas law equation, to solve for the concentration of $\mathrm{NO}\left(\mathrm{g}\right)$ .

$\mathrm{PV}=\mathrm{nRT}$

The unit of measurement is moles per litre. The ideal gas law equation can be rearranged to equal $\frac{\mathrm{n}}{\mathrm{V}}$.

$\frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}$

Solve for $\frac{\mathrm{n}}{\mathrm{V}}$ where $\mathrm{R}=0.0821\frac{\mathrm{L}\times \mathrm{atm}}{\mathrm{mol}\times \mathrm{K}}$.

$\begin{array}{rcl}\frac{\mathrm{n}}{\mathrm{V}}& =& \frac{\mathrm{RT}}{\mathrm{P}}\\ & =& \frac{\left(2.26\times {10}^2 \mathrm{atm}\right)}{\left(0.0821{\displaystyle \frac{\mathrm{L}\times \mathrm{atm}}{\mathrm{mol}\times \mathrm{K}}}\right)\left(1850 \mathrm{K}\right)}\\ \frac{\mathrm{n}}{\mathrm{V}}& =& 1.49\times {10}^{-4}\mathrm{mol}/\mathrm{L}\end{array}$

Convert the units from $\mathrm{mol}/\mathrm{L}$ to $\mathrm{\mu g}/\mathrm{L}$

$1.49\times {10}^{-4} \frac{\mathrm{mol}}{\mathrm{L}}\times \frac{30.01 \mathrm{gNO}}{1 \mathrm{molNO}}\times \frac{1 \mathrm{\mu g}}{1\times {10}^{-6} \mathrm{g}}=4.46\times {10}^3 \mathrm{\mu g}/\mathrm{L}$