Here we are asked to think about a reaction that occurs in the gas phase and his XO thermic for these five different conditions. How is the equilibrium constant? K impacted first, removing a reactor Well, we know from Lee Shot liaise principle that removing a reactant shifts the equilibrium of assistant itself. The equilibrium, constant itself K remains unchanged. That's because it's an equilibrium constant, and so it doesn't actually depend on the conditions you apply to it at a given time in terms of concentrations or, in this case, partial pressures of the gases in your system. Likewise, removing a product would invoke a leash Appy ace principle, but you still would see no change in K as it is an equilibrium constant. Similarly, if you decrease the volume of your system, we know that that would shift equilibrium towards the side that has fewer moles of gas. And so, while this might impact the exact composition of the system that you see the equilibrium, constant itself will not change. This is because equilibrium constance are only temperature dependent, not pressure, dependence. But they are temperature dependent. So now we have to think when we get to decreasing the temperature. We're told that the reaction is XO thermic. When we hear XO thermic, we can think of heat as a product if we add more heat than we're inhibiting this reaction from going forward, and so the reaction will actually shift back towards the starting materials. Therefore, the reaction constant K will decrease. Finally adding a catalyst and in a catalyst tends to speed things up. But ultimately there is no change in K since it's just a description of the ratios at equilibrium.