Question

Consider the decomposition equilibrium for dinitrogenpentoxide:

$\mathbf{2}{\mathit{N}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{5}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{4}\mathit{N}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

At a certain temperature and a total pressure of1.00 atm, the N₂O₅ is 0.50% decomposed (by moles)at equilibrium.

a. If the volume is increased by a factor of 10.0, willthe mole percent of N₂O₅ decomposed at equilibrium be greater than, less than, or equal to 0.50%?Explain your answer.

b. Calculate the mole percent of N₂O₅ that will bedecomposed at equilibrium if the volume isincreased by a factor of 10.0

Short answer

(a) If the volume is increased by a factor of 10, the mole percent of N₂O₅ decomposed at equilibrium will be greater than 0.50% as the equilibrium shifts towards right.

(b) 2 mole percent of N₂O₅ will be decomposed at equilibrium if the volume is increased by a factor of 10.

Step-by-step solution

Explanation regarding part (a)

(a)

If the volume is increased by a factor of 10, the mole percent of N₂O₅ decomposed at equilibrium will be greater than 0.50%, as the equilibrium shifts towards the right.

Determine the mole percent

(b)

It has been stated that at a certain temperature there is total pressure of 1.00 atm, and the N₂O₅ is 0.50% decomposed (by moles) at equilibrium.

As relation between pressure and numberof moles of an element are proportional to each other, therefore, it can be stated the change in atmospheric pressure of N₂O₅ will be 0.5%=0.005 atm. As per stoichiometric ratio change for NO₂ and O₂ are 0.010 and 0.0025 respectively.

$\begin{array}{l}2N_2{O}_5\rightleftharpoons 4N{O}_2+O_2\\ Initial:1atm00\\ Change:-0.005+0.010+0.0025\\ Equil:0.9950.0100.0025\end{array}$

Now the equilibrium constant will be

$\begin{array}{l}{K}_P=\frac{{\left[N{O}_2\right]}^4\left[O_2\right]}{{\left[N_2{O}_5\right]}^2}\\ K_P=\frac{{\left[0.01atm\right]}^4\left[0.0025atm\right]}{{\left[0.995atm\right]}^2}\\ K_P=2.52\times {10}^{-11}at{m}^3\end{array}$

It has been stated that the new volume is 10.0 times greater than the old volume. Therefore, partial pressure of N₂O₅ will decrease by a factor of 10.0 as pressure is

inversely proportional to the volume$P\infty \frac{1}{v}$

Therefore, pressure will be

$P_{N_2{O}_5}=1atm\times \frac{1}{10.0}=0.100atm$

Now let the change be x.

$\begin{array}{l}2N_2{O}_5\rightleftharpoons 4N{O}_2+O_2\\ Initial:0.1atm00\\ Change:-2x+4x+x\\ Equil:0.1-2x4xx\end{array}$

$\begin{array}{c}{K}_P=\frac{{\left[N{O}_2\right]}^4\left[O_2\right]}{{\left[N_2{O}_5\right]}^2}\\ K_P=\frac{{\left[4x\right]}^4\left[x\right]}{{[0.1-2x]}^2}\\ K_P=2.52\times {10}^{-11}at{m}^3\\ \frac{{\left[4x\right]}^4\left[x\right]}{{[0.1-2x]}^2}=2.52\times {10}^{-11}at{m}^3\\ 2x=2\times {10}^{-3}atm\end{array}$

P_N2O5 decomposed= 2×10^-3atm

As pressure and moles are directly proportional therefore mole percent decomposed will be

$\frac{2\times {10}^{-3}}{0.1}\times 100=2.0\%N_2{O}_5$