3
$\begingroup$

My chemistry textbook says that $K_{p}$ indicates the equilibrium position. It then says that $kp$ will not change if pressure is increased, but it also says that the position of equilibrium does change if pressure is increased.

Let me restate that: $K_{p}$ tells you the equilibrium position. Increasing pressure does not change $K_p$ (indicating the equilibrium position does not change) but does change the equilibrium position. This is clearly a contradiction, so how should I properly define equilibrium position and equilibrium constants? I've never really had a clear understanding of what equilibrium position means.

$\endgroup$
  • 1
    $\begingroup$ Mind your Ps and Qs -- or rather, mind your ks and Ks. A lowercase 'k' is usually used to denote a rate contstant, while an uppercase K is an equilibrium constant. $\endgroup$ – owjburnham Oct 26 '17 at 19:28
2
$\begingroup$

You're using the word "pressure" too casually. There are partial pressures of species and total pressure (which is the sum of all partial pressures of all species, regardless of if they're participating in the equilibrium or not).

Partial pressures of reactants and products factor into the equilibrium constant $K_{\mathrm{p}}$. If you increase the partial pressure of a gas that is not participating in the equilibrium, none of the other partial pressures change, so the equilibrium is unaffected, but the total pressure will change. Thus, depending on the circumstances, changing the total pressure may or may not change the position of equilibrium.

The key is you should focus only on the partial pressures of interest. All other pressures are irrelevant, unless they somehow tell you about a change in a partial pressure of interest.

$\endgroup$
1
$\begingroup$

Kp tells you the value of the partial pressure term at equilibrium. The partial pressure term itself may vary but at equilibrium, it is always the value of Kp.

The partial pressure term mentioned is the expression for Kp used to calculate Kp but the concentrations are not at equilibrium but at the point in time when you made the measurements.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.