gchem Qs

[TimeforDAT]

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1. what volume of water would be needed to dilute 50 ml of a 3M HCl solution to 1M?

2. the equilibrium of N2O4 (g) <--> 2NO2(g) delta H = 15 kcal/mol can be shifted to the right by: decreasing the temperature....someone explain this

 

[IsRaEl21]

1) Since you are diluting you need to use the Formula M1V1=M2V2. So we have 3M*50ml=1M*V2 by solving we get v2= 150. Therefore, the amount of water that is needed is 150ml-50ml= 100ml of pure water needed to dilute. Don't confuse the V2 with the pure amount of water that is needed.

2)Heat+ N2O4 (g) <--> 2NO2(g) since this is an endotheromic reaction we need to increase the temperature in order to shift the reaction foward to the right. Remember endotheromic reactions you need to increase temp. in order to drive the reaction foward. However, in exotheromic reactions you need to decrease temp. in order to drive the reaction foward. If you really want to know all of this material in depth, I suggest you to get the destroyer. Dr. Romano explaines it in a very easy and clear way, and there are many really good practice questions. I really hope I helped you. Good Luck!!

 

[whoaaitzkyle]

1. what volume of water would be needed to dilute 50 ml of a 3M HCl solution to 1M?

2. the equilibrium of N2O4 (g) <--> 2NO2(g) delta H = 15 kcal/mol can be shifted to the right by: decreasing the temperature....someone explain this

hey, ok so I'm not very good at gchem but I'll take a stab at it.

I know for #2 since the delta H is positive it means it's an endothermic reaction. By adding heat, due to le chatlier's principle (don't think I spelled that right but you know what I mean) the reaction would go to the right. Another way to look at it is....add heat to the reactant side since it's endothermic. Since you are adding more reactants, it will favor products.

As for #1 I always get confused which formula to use. At first instinct I would use M1V1=M2V2 but I'm not sure if that's correct. Hopefully someone else can correct me! Hope that helped!

 

[levyusupov]

get the DAT destroyer, all these concepts and others are explained in the best way so u can understand and grasp it. very recommended.

 

[TimeforDAT]

wait..i'm not getting something..you guys said for the second question that it is an endothermic reaction and that increasing temperature will drive the reaction to the right, the product side..i get all that.. but the question asks how you would shift the reaction to the right by DECREASING temp...what am i missing here?

 

[marscole]

i think you may apply Le Chatetlier's Rule by taking away 2NO2 (g). Then, the reaction will shift to the right when the temp is decreasing.

If adding pressure, i dont think it might work.

 

[TimeforDAT]

i think you may apply Le Chatetlier's Rule by taking away 2NO2 (g). Then, the reaction will shift to the right when the temp is decreasing.

If adding pressure, i dont think it might work.

the question doesn't say anything about taking away 2NO2...and even if that were the case, wouldn't the reaction only shift right with more heat, since heat is on the reactant side, product is gone, hence more heat means more product.

 

[IsRaEl21]

You want to drive the reaction foward in endotheromic reaction, than you need to add heat, it's only for endotheromic reactions. I really suggest you to get the destroyer b/c Dr. really explaines it well.

 

[porschex911x]

1. Agree with the above 100mL.

2. In the reaction you have N2O4 <---> 2 NO2. You know it is endothermic therefore the heat will be on the reactants side, and for exothermic reaction it will be on the right because you are releasing heat. Now by reducing temperature you are reducing the amount of heat on the reactants or the left side. A decrease in temp will cause a decrease in heat,To balance that loss of heat the reaction will drive to the left.

 

[TimeforDAT]

1. Agree with the above 100mL.

2. In the reaction you have N2O4 <---> 2 NO2. You know it is endothermic therefore the heat will be on the reactants side, and for exothermic reaction it will be on the right because you are releasing heat. Now by reducing temperature you are reducing the amount of heat on the reactants or the left side. A decrease in temp will cause a decrease in heat,To balance that loss of heat the reaction will drive to the left.

i get everything u guys are saying about adding heat to reactants and so forth, but it still doesn't answer the question as to why decreasing the temperature would drive the reaction to the right (Which is what the question wanted you to figure out).

ohh and by the way..the question is from kaplan, i know they make mistakes..could this possibly be a mistake on their part?

 

[bryanttttttta]

It doesn't, b/c it's shift to the LEFT for this endothermic reaction. Kaplan is probably wrong.