which side it goes?

[Dencology]

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hey guys, this problem is really bothering me: my ans: is C but it says that the ans: is A. According to Destroyer which when we have an exothermic rxn heat is given off, which means that heat is on the product side. Now, if we raise the temp, the reaction will move toward the reactant side. But how come in this question, the rxn moves toward the product


2CO (g) + O2 (g) -------> 2CO2 (g)
if a container holds only CO and O2, what effect will raising the temp have on the forward reaction:

a. the forward rxn rate will inc.
b. the forward rxn rate will remain the same.
c. the forward rxn rate will decrease.
d. the reaction will not go forward

. what is the rate constant dependent on?
a. temp
b. [reactant]
c. [product]
d. density

 

[DROCKINDAHOUSE]

think its endothermic

 

[tranv117]

hey guys, this problem is really bothering me: my ans: is C but it says that the ans: is A. According to Destroyer which when we have an exothermic rxn heat is given off, which means that heat is on the product side. Now, if we raise the temp, the reaction will move toward the reactant side. But how come in this question, the rxn moves toward the product


2CO (g) + O2 (g) -------> 2CO2 (g)
if a container holds only CO and O2, what effect will raising the temp have on the forward reaction:

a. the forward rxn rate will inc.
b. the forward rxn rate will remain the same.
c. the forward rxn rate will decrease.
d. the reaction will not go forward

The question says that you initially only start with CO and O2. So increasing the temperature will increase kinetic energy of the reactants leading to more collisions to form CO2. Now, had you had both reactants and products to start with then you would be correct that increasing temp would push the reaction left.

 

[Dencology]

guys i still don't understand the ans. would you explain in a different way?

 

[tranv117]

you dont have any CO2. Basically the reaction hasnt even taken place yet. Only have reactants. So whether or not the reaction is endothermic or exothermic, you dont know. BUT, if you add heat, you'll increase reactant collisions leading to CO2 formation. That's why the reaction must go forward when increasing the temp.

Doesn't really get any simpler then that. the key here is that you ONLY have CO and O2 gas

 

[Dencology]

ok. i got you now do you know if freezing water is endothermic or exothermic?
my guess is endothermic because water is gaining surrounding heat. but i have the answer as exothermic, since heat must be removed from water to make it freeze

what do you have to say?

 

[anesthesia_209]

ok. i got you now do you know if freezing water is endothermic or exothermic?
my guess is endothermic because water is gaining surrounding heat. but i have the answer as exothermic, since heat must be removed from water to make it freeze

what do you have to say?

It's exothermic.

To freeze water, you have to remove energy.
If you keep adding energy (endothermic), the water heats up and eventually becomes a gas.

 

[Doa110]

ok. i got you now do you know if freezing water is endothermic or exothermic?
my guess is endothermic because water is gaining surrounding heat. but i have the answer as exothermic, since heat must be removed from water to make it freeze

what do you have to say?

Look at the "Latent heat of diffusion thread". I have explained about freezing water. So are asking about the temp where the water is in the process of solidification? or when it is already frozen?
For the 1st case, it would be exothermic, but the latter must be endothermic if the environment has higher temp that the frozen water.

 

[joonkimdds]

The problem didn't tell us whether it's endothermic or exothermic so it's impossible to solve this problem.

 

[rdhdds1]

hey guys, this problem is really bothering me: my ans: is C but it says that the ans: is A. According to Destroyer which when we have an exothermic rxn heat is given off, which means that heat is on the product side. Now, if we raise the temp, the reaction will move toward the reactant side. But how come in this question, the rxn moves toward the product


2CO (g) + O2 (g) -------> 2CO2 (g)
if a container holds only CO and O2, what effect will raising the temp have on the forward reaction:

a. the forward rxn rate will inc.
b. the forward rxn rate will remain the same.
c. the forward rxn rate will decrease.
d. the reaction will not go forward

. what is the rate constant dependent on?
a. temp
b. [reactant]
c. [product]
d. density

One the first question you have 3 moles on the reactant side (2 CO and 1 O2) and you only have 2 on the product side ( 2 CO2).

3-------------> 2 + heat

the only direction that the reaction can go to is the right due to there being more space since there are only 2 moles on the product side. The system has to relieve the pressure that is on the reactant side by sliding to the product side.

I hope that helps.


For the second question it's exothermic.

"When a process occurs in which the system absorbs heat, the process is called endothermic (into). Melting of ice is an example. A process in which the system loses heat is called exothermic (out of). Freezing water is an example"

 

[Deontologist]

hey guys, this problem is really bothering me: my ans: is C but it says that the ans: is A. According to Destroyer which when we have an exothermic rxn heat is given off, which means that heat is on the product side. Now, if we raise the temp, the reaction will move toward the reactant side. But how come in this question, the rxn moves toward the product


2CO (g) + O2 (g) -------> 2CO2 (g)
if a container holds only CO and O2, what effect will raising the temp have on the forward reaction:

a. the forward rxn rate will inc.
b. the forward rxn rate will remain the same.
c. the forward rxn rate will decrease.
d. the reaction will not go forward

. what is the rate constant dependent on?
a. temp
b. [reactant]
c. [product]
d. density

I think exothermic and endothermic matters only when we want to study change in reaction that are at equilibrium. If the first reaction was in equilibrium we would have needed "thermicity" to get the answer, but since the reaction haven't started yet (i.e there is no product yet), giving in heat would shift the rxn to right.

Correct me if I am wrong.

 

[Isradoc1983]

ok i'll try to explain it differently. correct me if i am wrong.
If the correct answer is A then the reaction needs to be endothermic. The way i see it we're adding heat/energy to make new bonds( on the left) i.e make CO2 so it makes sense its endothermic.
idk about this explanation but thats how i'd justify it.
However, if the question says its endothermic it must be C

 

[tranv117]

you cannnot say if its endothermic or exothermic with this information unless you guys have the bond energies of CO2 and O2 and CO. Yes, we're forming bonds to make CO2 but you MUST realize that in the reactants you are also breaking bonds! So really, this has nothing to do with being endo or exothermic. It's just a matter of reactants and adding heat it order to increase effective collisions to form new bonds making CO2.

 

[ZenoVT]

This is combustion reaction and therefore is exothermic. The key word is that a container holds only CO and O2 so the only way to go is forward. So you do not use Le Chatelier's to answer the question and use thermodynamics. Increasing temperature will increase the rates where CO and O2 collide and react, and temperature is also what affects the rate constant.

 

[tranv117]

This is combustion reaction and therefore is exothermic. The key word is that a container holds only CO and O2 so the only way to go is forward. So you do not use Le Chatelier's to answer the question and use thermodynamics. Increasing temperature will increase the rates where CO and O2 collide and react, and temperature is also what affects the rate constant.

thanks