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Calculate the value of $K_p$ for the following equation:

$$\ce{PCl3 (g) + Cl2 (g) <=> PCl5 (g)} ,\qquad \mathrm{R = 0.0821 \frac{L\times atm}{mol\times K}}$$

This question appeared on my test. I know how to find $K_p$ if I was given $K_c$ $(K_{\mathrm{p}} = K_\mathrm{c} \times (\mathrm{R} \times T)^{n})$ or if I was given the partial pressures of all of the parts at equilibrium, but none of those values were given; only the equation. My professor marked the answer as $0.041$ (no units), but I have no idea how to calculate that if I wasn't given any values.

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  • $\begingroup$ Notice that I replaced \mathrm with \ce for your chemical equation. Just like that, it properly rendered the subscripts and reaction arrows! $\endgroup$ – airhuff Feb 21 '17 at 3:19
  • $\begingroup$ Interesting that the answer is 1/2 of R, and that there 1/2 as many moles of gas on the right side of the equation. $\endgroup$ – airhuff Feb 21 '17 at 3:45
  • $\begingroup$ Thanks for the help! Interesting observation. I wonder if my professor accidentally took the moles of each side and divided them (products/reactants) instead of using the partial pressures. $\endgroup$ – Eric Feb 21 '17 at 4:03
  • $\begingroup$ Were you given any other information at all for the problem? I think the key is that the total pressure of the products is half of the total pressure of the reactants if the volume is constant. $\endgroup$ – airhuff Feb 21 '17 at 4:12
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    $\begingroup$ Were you allowed to use tables of free energies of formation or enthalpies and entropies of formation? $\endgroup$ – Chet Miller Feb 21 '17 at 13:09

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