Question

An atmospheric chemist fills a container with gaseous ${\mathit{N}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{5}}$ to a pressure of 125 kPa, and the gas decomposes to$\mathit{N}{\mathit{O}}_{\mathbf{2}}$ and ${\mathit{O}}_{\mathbf{2}}$. What is the partial pressure of ${\mathit{P}}_{\mathbf{N}{\mathbf{O}}_{\mathbf{2}}}$, (in kPa), when the total pressure is 178 kPa?

Short answer

The partial pressure of$N{O}_2$ is equal to 71 kPa when the total pressure is 178 kPa.

Step-by-step solution

Step 1: The chemical equation for the decomposition

Write the chemical equation for the decomposition of :

$N_2{O}_{5\left(g\right)}\to N{O}_{2\left(g\right)}+O_{2\left(g\right)}$

Balance the nitrogen atoms:

$N_2{O}_{5\left(g\right)}\to 2N{O}_{2\left(g\right)}+O_{2\left(g\right)}$

Balance the oxygen atoms:

$N_2{O}_{5\left(g\right)}\to 2N{O}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}$

As a result, one mole of $N_2{O}_5$breakdown produces two moles of $N{O}_2$and half a mole of$O_2$ . As a result, the number of moles in the goods increases.

Step 2: Determine the partial pressure of NO2

A gas's pressure is proportional to its number of moles, and a change in the number of moles signals a change in reaction pressure. Consider x to be the drop in $N_2{O}_5$pressure. Relate the pressure change in terms of the starting pressure, $P_{initial}$, and the final pressure, $P_{final}$, as follows:

$$\begin{gathered} N_2{O}_{5\left(g\right)}\to 2N{O}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)} \\ {P}_{initial}12500 \\ {P}_{final}\text{125 - x}2x\frac{1}{2}x \end{gathered}$$

$$\begin{gathered} {\text{TotalP}}_{\text{final}}\text{=}\left(\text{125 - x}\right)\text{+ 2x +}\frac{\text{1}}{\text{2}}\text{x} \\ \text{178kPa =}\left(\text{125 - x}\right)\text{+ 2x +}\frac{\text{1}}{\text{2}}\text{x} \\ \text{x = 35.33 kPa} \end{gathered}$$

Determine the partial pressure of$N{O}_2$:

$$\begin{gathered} P_{N{O}_2}=2x \\ =2\times 35.33kPa \\ =71kPa \end{gathered}$$

Thus, the partial pressure of$N{O}_2$ is equal to 71 kPa.