# Chem question

#### dental13

10+ Year Member
7+ Year Member
Okay I have attached a Gchem question from a past DAT.

The answer is 3. Yes I do know that this will shift the equilibrium to the right. But sine this is endothermic because of +H, therefore increasing temperature will also shift the reaction to the right, which is the anser choice 2.

is the question just incorrect? or I am getting confused with basci stuff.. #### Attachments

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#### SN1

Removed
Okay I have attached a Gchem question from a past DAT.

The answer is 3. Yes I do know that this will shift the equilibrium to the right. But sine this is endothermic because of +H, therefore increasing temperature will also shift the reaction to the right, which is the anser choice 2.

is the question just incorrect? or I am getting confused with basci stuff.. Increasing the pressure will shift the equilibrium to the direction that makes less moles of gas so answer 3 is correct.

if the delta H listed on that problem is indeed positive which means it is an endothermic reaction and YES you are right that increasing the temperature will also shift the reaction to the right.

Honestly I think answer 2 should also be correct. Unless "keeping the pressure constant has something to do with it"

When you increase the temperature while keeping the pressure constant, you are increasing the volume which means that you are decreasing the pressure and that would shift equilibrium to the right as well making more moles of gas.

so yeah, must be a typo or something

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#### premac

10+ Year Member
7+ Year Member
I believe that you guys misunderstood the given problem.
+ 22 Kcal seems to be a PART of the whole equation, not a separate information.
If it was a side information, you could say that the equation is endothermic, since the value is positive.
However, since it's part of the equation, the problem is saying that heat is in the "product" side of the equation.
Thus, this would be an exothermic equation. Being exothermic, increasing temperature would shift the equilibrium to the left.

#### ijk90825

7+ Year Member
it's N2 + 3H2 -> 2NH3 + heat, so it's exothermic not endothermic. heat is the end product.
catalyst wont shift the equilibrium, increase in temperature will shift toward left, increase pressure will shift it to the right and take NH3 out will shift it to the left. hope it makes sense

#### SN1

Removed
I believe that you guys misunderstood the given problem.
+ 22 Kcal seems to be a PART of the whole equation, not a separate information.
If it was a side information, you could say that the equation is endothermic, since the value is positive.
However, since it's part of the equation, the problem is saying that heat is in the "product" side of the equation.
Thus, this would be an exothermic equation. Being exothermic, increasing temperature would shift the equilibrium to the left.
Yeah you are absolutely correct. When I saw the word +22kcal, I automatically assumed that was the value of deltaH lol. I didn't even notice that. So yes the heat is on the product side and is therefore exothermic.

Thanks for clarifying

OP
D

#### dental13

10+ Year Member
7+ Year Member
it's N2 + 3H2 -> 2NH3 + heat, so it's exothermic not endothermic. heat is the end product.
catalyst wont shift the equilibrium, increase in temperature will shift toward left, increase pressure will shift it to the right and take NH3 out will shift it to the left. hope it makes sense
Yes, it is the part of the system.
Thank you 