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My work:

${K_p = 28.4 = \frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{[NOBr]^2}{[107]^2[160]}} \rightarrow$$\ce{[NOBr]=7212 torr}$.

This doesn't match up with the given answer. I don't see why 262 torr is the correct answer. I correctly set up the equilibrium constant expression and everything in the reaction equation is in the gas phase so they are all included and I used only equilibrium concentrations.

Another problem with the same issue:

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  • $\begingroup$ Weird. You math is right. If the constant is based on mole fraction then I get 19.8 for the constant based on the answer of 262 torr. $\endgroup$ – MaxW Feb 18 '16 at 3:39
  • $\begingroup$ It should be products over reactants, but inverted: $$ \dfrac{(107^2) (160)}{262^2} = 26.7 $$ It seems problem/answer just messed up. $\endgroup$ – MaxW Feb 18 '16 at 4:18
  • $\begingroup$ @MaxW this is a consistent issue. See my edit. I found that I get the correct answer after converting all the torr values to atm and keeping Kp as is. Why is it that I must convert torr to atm? $\endgroup$ – Dissenter Feb 18 '16 at 14:26
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In general $k_{p}$ is not a dimensionless quantity. In the case of the reaction $$\ce{2NO + Br2 <=> 2NOBr}$$ the unit of measurement for $k_{p}$ is in fact $$\frac{pressure^2}{pressure^2 * pressure} = pressure^{-1}.$$ Thus the value 28.4 of the equilibrium constant given in the assignment is incomplete.

Given the correct answer, it turns out that $k_p$ is actually given in $\mathrm{atm^{-1}}$. Since $$\mathrm{28.4~atm^{-1}} = 28.4*(\mathrm{760~Torr})^{-1} = 0.0374~\mathrm{Torr}^{-1}$$ you should use this value for $k_p$ if you prefer to do your calculation with $\mathrm{Torr}$.

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  • $\begingroup$ how do we know the units a priori $\endgroup$ – Dissenter Feb 20 '16 at 4:47
  • $\begingroup$ The unit must be given along with the numerical value. The exponent can also be derived from the stoichiometric coefficients as shown above. In the special case that the exponent is zero the unit of measurement is 1, i.e. neither Pa, nor atm, nor Torr, or any other unit of pressure measurement. $\endgroup$ – aventurin Feb 20 '16 at 16:46

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