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How Does Kp Change With Temperature



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Question
What is the effect of temperature on Kp and Kc values?
I am getting confused with when the position of equlibrium shifts to the right/left and when it is an exothermic/endothermic reaction and how does Kc and Kp values change with temperature...

Answer
N2O4 === 2NO2 -+58.0 The foreward reaction is endothermic. (it takes in heat from the surrroundings). Therefore if you heat it up, the equilibrium will shift towards the right in order to counteract this change (eg to cool the temperature down it takes heat from the surroundings). So heating up means the forward reaction is favoured. Equilibrium shifts to right. Cooling down means the backward reaction is favoured. Equilibrium shifts to left. Kc and Kp i think are unaffected by temperature because they are in proportion to the concentration of the different reactants and products. NOT the temperature at which these are.



Question
Why do the constants of equilibrium Kc and Kp only change with temperature?
Is this just proven using empirical evidence, or is there some logical explanation? Why don't changing pressures and concentration change the constants of equilibrium?

Answer
There are two ways to look at this. The concept of equilibrium and the equilibrium expressions come from the law of mass action. The law of mass action at a given temperature will always generate the same ratio of concentrations of products to reactants regardless of the starting concentrations of either. This final ratio is the value of Keq. A second way is to look at from a kinetics standpoint. At equilibrium the forward rate of reaction is equal to the reverse rate of reaction. At equilibrium, the only fact that will affect the ratio of the rates of the forward and reverse reactions is temperature.



Question
Why changes in temperature affect the value or the equilibrium constant?
Just wondering, im curious why changing the pressure or concentration, the system automatically shifts one way or another to maintain the equilibrium balance but temperature doesnt?

Answer
Le Chatelier's Principle is a very useful tool to predict the direction of the equilibrium. There are mainly 3 factors which disturbs the equilibrium. 1) Change in concentration 2) Change in pressure (by changing volume) 3) Change in temperature I will explain the effects of these changes by using Le Chatelier's Principle. The application is rather simple. If something increases, try to decrease it. If something decreases, try to increase it. I will use the following reaction to explain all these effects. 2SO2(g) + O2(g) ↔ 2SO3(g) .......... ΔH = −197 kJ 1) Change in concentration: - If more O2 is added to the system at equilibrium, the system will try to decrease the added O2. Therefore the equilibrium shifts to right. That is forward reaction will be effective. SO2 and O2 will react and will produce more SO3. - Same thing occurs when more SO2 is added. - If more SO3 is added. the system will try to decrease the added SO3. Therefore the equilibrium shifts to left. That is reverse reaction will be effective. SO3 will decompose into SO2 and O2. - If some SO3 is removed from the system at equilibrium, the system will try to increase the removed SO3. Therefore the equilibrium shifts to right. That is forward reaction will be effective. SO2 and O2 will react to form SO3. - If some SO2 is removed from the system at equilibrium, the system will try to increase the removed SO2. Therefore the equilibrium shifts to left. That is reverse reaction will be effective. SO3 will decompose into SO2 and O2. 2) Change in pressure (by changing volume) - If the pressure of the system is increased by decreasing the volume, the system will try to decrease the pressure. Since the partial pressures of gases are directly proportional to their number of moles, the equilibrium shifts to right. Why? Because, at the right there are 2 moles, at the left there are 4 moles. As the number of moles decreases, the pressure also decreases. - If the pressure of the system is decreased by enlarging the volume, the system will try to increase the pressure. The reverse of the first case will happen. - Same thing occurs when some O2 is removed. NOW YOUR QUESTION; 3) Change in temperature - If the temperature of the system is increased, the system will try to decrease this increase by consuming heat. Since the forward reaction is exothermic (ΔH = −197 kJ), the reverse reaction will be effective. The equilibrium will shift to left. Endothermic reverse reaction will consume heat. - If the temperature of the system is decreased, the system will try to increase this decrease by poducing heat. Since the forward reaction is exothermic (ΔH = −197 kJ), the forward reaction will be effective. The equilibrium will shift to right. Exothermic forward reaction will produce heat. This is IMPORTANT for you; In cases of concentration or pressure changes, the system will readjust itself to satisfy Kc or Kp value. But, in case of temperature changes, the value of Kc or Kp changes. Whenever equilibrium shifts to right by the effect of temperature change, Kc and Kp will be greater and whenever equilibrium shifts to left by the effect of temperature change, Kc and Kp will be smaller. 4) Effect of Catalyst: Catalysts increase the rates of both the forward and reverse reactions equally. Thus, they reduce the time to reach the equilibrium. They have no effect on either the yield of the reaction or the equilibrium constants.




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