We’re being asked to **determine the pressure of NO _{2}(g) at equilibrium for the following reaction**:

2 NO_{2}(g) ⇌ 2 NO(g) + O_{2}(g) K_{p} = 158

Recall that the ** equilibrium constant** is the ratio of the products and reactants.

We use **K _{p}** when dealing with pressure and

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}\frac{{\mathbf{P}}_{\mathbf{products}}}{{\mathbf{P}}_{\mathbf{reactants}}}}$ $\overline{){{\mathbf{K}}}_{{\mathbf{c}}}{\mathbf{=}}\frac{\mathbf{\left[}\mathbf{products}\mathbf{\right]}}{\mathbf{\left[}\mathbf{reactants}\mathbf{\right]}}}$

*Note that solid and liquid compounds are ignored in the equilibrium expression.*

A sample of pure NO_{2} gas heated to 1000 K decomposes:

2NO_{2}(g) ⇌ 2NOg) + O_{2}(g) K_{p} = 158

Analysis shows that the partial pressure of O_{2} is 0.25 atm at equilibrium. Calculate the pressure of NO_{2}(g) at equilibrium.

A. 0.090 atm

B. 0.015 atm

C. 0.020 atm

D. 0.038 atm

E. 0.075 atm

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