Which of the folowing rate of reaction at equillibrium depends upon?1-temp. 2-pressure,3-catalyst?

I don't belive i am doing this but i can prove i am right. First of all equilibrium of a temperature means constant of a temperature at a suitable degree while an exprement is done. >Van't Hoff-Boyle's law:If behavior of dissolved molecules is considered to be analogous to gas molecules,then the osmotic pressure could be considered to be due to bombardment of solute particles or the concentration of solute at CONSTANT TEMPERATURE. Thus,Van't Hoff-Boyle's law is stated as,at CONSTANT TEMPERATURE the osmotic pressure (pi)(which is 3.14) of a dilute solution is directly propotional to its molar concentration or inversly proportinal to the volume of the solution (pi) direct proptional(in symbolic form) C (C is concentration in mol L^-1)...(Constant T) If n moles of solute is dissolve in V litres then, C = n/V Hence, (pi) direct propotionl(sym) n/V or (pi)direct propotion(sym) 1/v for a constant n Hence, (pi)V=constant or (pi)/C = constant. >Van 't Hoff - Charle's law The law stated as,the concentration remaining constant;the osmotic pressure of a dilute solution directly proportional to the absolute temperature. (pi) directly proportional to T (temp) i.e. pi/T = constant Also which is (pi) 1/T = constant. This is the relationship is similar to Charle's law for gases and known as van't Hoff-Charle's law for solution. If u are still confused about it then i am giving an easy example, Temperature have to be maintain in equilibrium will doing an exprement because of temperature change a solution can show the error in reading by being cooler(freezing) or hoter(melting) due to the change in room temperature.So such exprements are carried out in lab which is first made to be at Constant temperature. Plz, don't say me u didn't understand me(i have refered from the book of which i had give exam). & even if i haved proved it wrong,i can say u that other obtion u gave has much less possiblity as temperature does. Hope this will help u.