Question

Consider the following chemical equation:

$\mathbf{2}\mathit{N}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }{\mathit{N}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

If 25.0mL of NO₂ gas is completely converted to N₂O₄ gas under the same conditions, what volume will theN₂O₄ occupy?

Short answer

N₂O₄ will occupy 12.5mL volume.

Step-by-step solution

Information given in the question

No. of moles of N₂O (${\mathit{n}}_{N_{2}O}$) = 2mol

No. of moles of N₂O₄ (${n_{N_{2}O}}_4$) = 1mol

Volume of N₂O (${\mathit{V}}_{N_{2}O}$) = 25.0mL

Avogadro’s law

According to this law, the volume of a gas is directly proportional to its quantity (or, the no. of moles) at a constant temperature and pressure.

We can write it as,

$\mathit{V}\mathbf{}\mathbf{\infty }\mathbf{}\mathit{n}$ (at constant T and P)

Calculation of volume (VN2O4) of N2O4

From Avogadro’s law, we have,

$\begin{array}{rcl}\frac{{\mathbf{V}}_{{\mathbf{N}}_{\mathbf{2}}\mathbf{O}}}{{\mathbf{n}}_{{\mathbf{N}}_{\mathbf{2}}\mathbf{O}}}& \mathbf{=}& \frac{{\mathbf{V}}_{{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}}{{\mathbf{n}}_{{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}}\\ {\mathit{V}}_{{\mathbf{N}}_{\mathbf{2}}\mathbf{O}}& \mathbf{=}& \frac{{\mathbf{V}}_{{\mathbf{N}}_{\mathbf{2}}\mathbf{O}}\mathbf{\times }{\mathbf{n}}_{{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}}{{\mathbf{n}}_{{\mathbf{N}}_{\mathbf{2}}\mathbf{O}}}\end{array}$

$({\mathit{V}}_{N_{2}O}=25.0mL;{\mathit{n}}_{N_{2}O}=2mol;{{\mathit{n}}_{N_{2}O}}_4=1mol)$

Now,

$$\begin{gathered} {\mathit{V}}_{{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}\mathbf{=}\frac{\mathbf{25}\mathbf{\times }\mathbf{1}}{\mathbf{2}} \\ {\mathit{V}}_{{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{5} \end{gathered}$$

Hence, the volume of N₂O₄ is 12.5mL.

Therefore, N₂O₄ will occupy 12.5mL volume.