**N _{2}O_{4}(g) ⇌ 2NO_{2}(g) K_{p} = 0.25**

**Step 1:**** **Construct an ICE chart for the equilibrium reaction.

**Step 2: **Calculate the change (x) in the reaction using the equilibrium constant.

**N _{2}O_{4}(g) ⇌ 2NO_{2}(g) K_{p} = 0.25**

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}\phantom{\rule{0ex}{0ex}}\mathbf{K}\mathbf{=}\frac{{\mathbf{\left[}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\right]}}^{\mathbf{2}}}{\mathbf{\left[}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\right]}}\phantom{\rule{0ex}{0ex}}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{=}\frac{{\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{)}}^{\mathbf{2}}}{\mathbf{x}}\phantom{\rule{0ex}{0ex}}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{x}\mathbf{=}\mathbf{81}\mathbf{-}\mathbf{36}\mathbf{x}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{0}\mathbf{=}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{36}\mathbf{.}\mathbf{25}\mathbf{x}\mathbf{+}\mathbf{81}}$

Now, we will use the ** quadratic formula** to solve for x. The quadratic formula is:

At a particular temperature, *K*_{p} = 0.25 for the reaction

N_{2}O_{4}(*g*) ⇌ 2NO_{2}(*g*)

b. A flask containing only NO_{2} at an initial pressure of 9.0 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases.

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