We are asked to calculate the concentrations of N_{2}O_{4} and NO_{2} when this reaction reaches equilibrium.

Step 1. Calculate molarity N_{2}O_{4}

$\overline{){\mathbf{M}}{\mathbf{}}{{\mathbf{N}}}_{{\mathbf{2}}}{{\mathbf{O}}}_{{\mathbf{4}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mol}}{\mathbf{L}}}\phantom{\rule{0ex}{0ex}}\mathbf{M}\mathbf{}{\mathbf{N}}_{{\mathbf{2}}}{\mathbf{O}}_{{\mathbf{4}}}\mathbf{}\mathbf{=}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}{\mathbf{10}\mathbf{}\mathbf{L}}$

**M N _{2}O_{4} = 0.1 M**

Step 2. Set up ICE chart

Step 3. Calculate molarity in equilibrium

$\overline{){\mathbf{K}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{{\mathbf{\left[}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\right]}}^{\mathbf{2}}}{\left[{N}_{2}{O}_{4}\right]}}\phantom{\rule{0ex}{0ex}}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{\left[}\mathbf{2}\mathbf{x}\mathbf{\right]}}^{\mathbf{2}}}{[0.1-x]}$

x is negligible in the denominator since K is very small.

At a particular temperature, *K* = 4.0 x 10 ^{-7} for the reaction

N_{2}O_{4}(*g*) ⇌ 2NO_{2}(*g*)

In an experiment, 1.0 mole of N_{2}O_{4} is placed in a 10.0-L vessel. Calculate the concentrations of N_{2}O_{4} and NO_{2} when this reaction reaches equilibrium.

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the ICE Chart concept. You can view video lessons to learn ICE Chart. Or if you need more ICE Chart practice, you can also practice ICE Chart practice problems.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.