asdasdKp can be calculated as:

$\overline{){\mathbf{Kp}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{products}}{\mathbf{reactants}}}\phantom{\rule{0ex}{0ex}}\mathbf{Kp}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{p}}_{{\mathbf{NOBr}}^{\mathbf{2}}}}{{{\mathbf{p}}_{\mathbf{NO}}}^{\mathbf{2}}\mathbf{}\mathbf{\times}\mathbf{}{\mathbf{p}}_{{\mathbf{Br}}_{\mathbf{2}}}}\phantom{\rule{0ex}{0ex}}\mathbf{28}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{p}}_{{\mathbf{NOBr}}^{\mathbf{2}}}}{{\mathbf{107}}^{\mathbf{2}}\mathbf{\times}\mathbf{}\mathbf{166}}\phantom{\rule{0ex}{0ex}}{\mathbf{p}}_{\mathbf{NOBr}}\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{28}\mathbf{.}\mathbf{4}({107}^{2}\times 166)}$

Consider the following reaction:

2NO(g) + Br_{2}(g) ⇌ 2NOBr(g)

Kp= 28.4 at 298 K

In a reaction mixture at equilibrium, the partial pressure of NO is 107 torr and that of Br_{2} is 166 torr.

What is the partial pressure of NOBr in this mixture?

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