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The second-order reaction A + B --------> C with first-orderin each reactant can be approximated as a pseudo-first-orderreaction ifa) A is present in large exces...

Question

The second-order reaction A + B --------> C with first-orderin each reactant can be approximated as a pseudo-first-orderreaction ifa) A is present in large excessb) At equilibriumc) In the presence of a catalystd) At high temperaturee) At low temperature

The second-order reaction A + B --------> C with first-order in each reactant can be approximated as a pseudo-first-order reaction if a) A is present in large excess b) At equilibrium c) In the presence of a catalyst d) At high temperature e) At low temperature



Answers

The rate of a reaction can be affected by
(A) The phase of the reactants.
(B) The temperature of the reactants.
(C) The presence of a catalyst.
(D) All of the above.
(E) None of the above.

So we're told that we have a reaction. A plus B. He goes to seed and we're told that this is a single step. And so if this is a single step, that means A and B have to be reacting. It's told that the Bible likely reaction. So all that tells us this should be a second order reaction. And so, you know, that's because if you have a one step reaction, all the molecules here, all the reactions have to collide one to second order. And that tells us then that the rate locks should be ray equals k times a concentration of a times a concentration of being. So this now this we can use this information toe, figure out all the answers. So a here that the half life equals Ellen of two over K. This is for a first order reaction. And so this doesn't This doesn't work. This hasn't cracked be about the rate of disappearance. This is just so I can actually located. This is wrong. C is gonna be appearing twice as fast as a is disappearing because the coefficients so be is wrong. See, we just said this is gonna be the rate because the Raila Haas to reflect that this is a single step and a and B are colliding in each of them. And so that's Ah, we know that c is correct and D is incorrect because that's the first order Integrated rate law and where have a second order reaction, so that could just not be true.

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.

There are several factors that increase or decrease the rate of a reaction and they all are associated with the collision theory. If we were to increase the concentration of a reactant, we would increase the rate at which they collide. If we increase the rate at which they collide, we increase the rate of the reaction. If we increase the temperature, we will also increase the rate at which they collide, and also the energy associated with the collision. Increasing both the collision rate and the energy of a collision increases the rate of a reaction. Any time we add a catalyst, whether it's a metal catalyst or an acid catalyst or whatever, a catalyst always lowers the activation energy, lowering the activation energy requires less energy for a collision and an increase in the rate of the reaction. So just the opposite. If we were to decrease the concentration, we would decrease the rate of collision and decrease the rate of the reaction. If we were to decrease the temperature, we would decrease the energy of the collision, decrease the frequency of a collision, and also decrease the rate of the reaction. But adding a catalyst again, a metal catalyst and acid catalyst, it doesn't matter that's always going to decrease the activation energy and in turn, increase the rate. So this would not decrease the rate of the reaction. It would actually increase it.

A catalyst has various effects on the chemical reaction in order to speed up the chemical reaction. This question wants you to answer the effects of the catalysts on various processes or properties associated with a chemical reaction. First is activation energy. Hopefully, you understand now that a catalyst allows for an increase in the rate of a chemical reaction by decreasing the activation energy. So this would be the effect of a catalyst on activation energy. The next one is in reference to the reaction mechanism. Often it achieves a decrease in activation energy by providing an alternate mechanism. So typically, a catalyst will alter the mechanism of the chemical reaction. However, the change in energy between the reactant and the products will be exactly the same. Delta H for the reaction will be constant whether you have a catalyst or not. Then, for Part D, it wants to know what happens to the rate of the forward reaction and for e the rate of the reverse reaction. Well, the rate of the forward reaction will increase. That's what we hope it will do with the addition of a catalyst. But in turn, when the activation energy in the forward direction is decreased, the activation energy in the reverse direction is also decreased. The height of this hill in the reverse direction is decreased and the height of the hill in the forward direction is decreased. So we also decrease the activation energy in the reverse direction and in turn increased the rate of the reaction in the reverse direction also.


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