New answers tagged


$K_p$ will not change with pressure, only with temperature. $Q_p$ will instantaneously increase upon applying the pressure for this reaction (why? see below), and will drive the reaction to reactants to re-achieve equilibrium (decreasing $Q_p$) until $Q_p$ = $K_p$ once again. From a LeChatelier approach, you have more gas molecules in the products, so an ...

Top 50 recent answers are included