Question

At ${\mathbf{25}}^{\mathbf{\circ }}\mathit{C}\mathbf{,}\mathbf{}{\mathit{K}}_{\mathbf{P}}\mathbf{\approx }\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}$for the reaction

${\mathit{N}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}\mathit{N}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

a. Calculate the concentration of NO (in molecules/cm³) that can exist in equilibrium in air atrole="math" localid="1657791556979" ${\mathbf{25}}^{\mathbf{\circ }}\mathit{C}$ . In air P_N2 = 0.8 atm and P_O2 = 0.2 atm.

b. Typical concentrations of NO in relatively pristine environments range from 10⁸ molecules/cm³ to 10¹⁰ molecules/cm³. Why is there a discrepancy between these values and your answer to part a?

Short answer

a. The concentration of NO in air is found to be $2\times {10}^{3}molecules/c{m}^3$.

b. The concentration of NO in the atmosphere is higher than the calculated concentration from the equilibrium constant, K value, because NO does not return to N₂ and O₂ once it reaches a more normal temperature environment.

Step-by-step solution

Calculate the partial pressure of NO.

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

$K_p=\frac{{P_{NO}}^2}{\left(P_{N_2}\right)\left(P_{O_2}\right)}$ …(1)

Substitute the value of K_p and the partial pressures of nitrogen and oxygen into Equation (1).

$1\times {10}^{-31}=\frac{{P_{NO}}^2}{\left(0.8\right)\left(0.2\right)}$

Solve for P_NO .

$P_{NO}=1\times {10}^{-16}atm$

Hence, the partial pressure of NO is $1\times {10}^{-16}atm$.

Calculate the concentration of NO in air.

The concentration of NO can be calculated by the following expression as follows:

$\left[NO\right]=\frac{P_{NO}}{RT}$ …(2)

Here,

[NO] is the concentration of NO.

P_NO is the partial pressure of NO.

R is the gas constant.

T is the absolute temperature.

Substitute as follows:

$\begin{array}{rcl}\left[NO\right]& =& \frac{1\times {10}^{-16}atm}{0.08206Latm/Kmol\times 298k}\\ & =& 4\times {10}^{-18}mol/L\\ & =& \frac{4\times {10}^{-18}mol/L}{1L}\times \frac{1L}{{10}^{3}c{m}^3}\sin ce1L={10}^{3}c{m}^3\\ & =& 4\times {10}^{-21}mol/c{m}^3\end{array}$

Since 1 mole of NO contains $6.022\times {10}^{23}molecules$.

$\frac{4\times {10}^{-21}mol}{1c{m}^3}\times \frac{6.022\times {10}^{23}molecules}{1mol}=2\times {10}^{3}molecules/c{m}^3$

Hence, the concentration of NO in air is found to be $2\times 1^{3}molecules/c{m}^3$.

Find a discrepancy between the original value and the calculated one.

The concentration of NO in the atmosphere is higher than the calculated concentration from the equilibrium constant, K value.

$$\begin{gathered} N_2\left(g\right)+O_2\left(g\right)\to 2NO\left(g\right)\dots \left(3\right) \\ 2NO\left(g\right)\to N_2\left(g\right)+O_2\left(g\right)\dots \left(4\right) \end{gathered}$$

Both Reactions (3) and (4) occur at such a slow rate that they are effectively nil. The activation energy is quite large, and very strong bonds must be broken.As a result, the reaction does not take place at low temperatures. Because the generation of nitric oxide is endothermic, it may be created in high-energy or high-temperature settings. Lightning produces some NO in nature, and vehicles are the largest artificial source. At these high temperatures, K increases, and the reaction rates rise as well, resulting in more NO generation. Because of the relatively slow rate, NO does not return to N₂ and O₂ once it reaches a more normal temperature environment.