Will this reaction cause an increase or decrease in temperature?

[ElectroMaster88]

Homework Statement

0.8 bar of the gas N2O4 and 0.2 bar of the gas NO2 are put inside a container with constant volume. The gases react according to the reaction N2O4 -> 2NO2 until they reach equilibrium. If we suppose that the temperature can change, will is increase or decrease because of the reaction? Assume that the initial temperature is 298K.

Relevant Equations

ΔH=ΔU+VΔP+PΔV
ΔU=ΔQ+ΔW

I have an idea for a solution, I just want to make sure if it makes sense of if it needs to be a bit more polished.

In a constant volume (i.e. ΔV=0) we have:
ΔH=ΔU+VΔP
ΔU=ΔQ

So we can get:
ΔH=ΔQ+VΔP
ΔQ=ΔH-VΔP

Now I can calculate ΔH using data about these gases, and I can calculate the volume using the ideal gas law, and then I can calculate VΔP, and then show that VΔP is much smaller than ΔH, so the sign of ΔH is the same as the sign of ΔQ.

So if ΔH is positive, then ΔQ is also positive, which means that the reaction is endothermic.
If the reaction is endothermic, it takes heat from the surroundings, so the temperature of the surroundings decreases.

And since the temperature of the gas is determined by the temperature of the surroundings, the answer is that the temperature will decrease.

Is my solution good? Does it need to be a bit refined? Is there a better solution?

 

[Mayhem]

You could also argue that you start with an excess of N2O2 ("dimer" of NO2) which will force the reaction right (to form NO2). Since bond breaking is endothermic, so must the overall process also be, assuming no other factors.

 

[ElectroMaster88]

That's an interesting way to think about it, but is my way also good?

And if the reaction is endothermic, does the temperature really decrease like I wrote?

 

[Chestermiller]

Homework Statement: 0.8 bar of the gas N2O4 and 0.2 bar of the gas NO2 are put inside a container with constant volume. The gases react according to the reaction N2O4 -> 2NO2 until they reach equilibrium. If we suppose that the temperature can change, will is increase or decrease because of the reaction? Assume that the initial temperature is 298K.
Relevant Equations: ΔH=ΔU+VΔP+PΔV

This equation is incorrect. It should read $\Delta H=\Delta U +\Delta (PV)$.

The correct way to analyze this problem is to recognize that,
for a constant volume insulated reactor, $\Delta U=0$. For the reaction proceeding from the initial standard state to the final standard state at constant volume and constant temperature , we have $$\Delta U^0=\Delta H^0-\Delta nRT^0$$ where $\Delta n=+1$. So, $$\Delta U^0=\Delta H^0-RT^0$$This is the amount of heat that would have to be added to the reactor (per mole of $N_20_4$ reacted) to hold the temperature of the reactor at 298K. But, the products will not be at 298 K, because the reactor is insulated. So we must have that $$\Delta U=\Delta m\Delta U^0+MC_v(T-298)=0$$ where $\Delta m$ is the number of moles of $N_2O_4$ which react, M is the final number of moles of the two gases in the reactor, and $C_V$, is the molar average heat capacity of the final mixture. Please note that T will have to be determined implicitly, since the equilibrium constant for the reaction changes with temperature.

 

[Mayhem]

That's an interesting way to think about it, but is my way also good?

And if the reaction is endothermic, does the temperature really decrease like I wrote?

Endothermic -> absorbs heat
Exothermic -> expels heat

The loss/gain of to the surroundings is then measured as a temperature decrease/increase. In bond-making (endothermic reaction), the majority of absorbed heat is stored as potential energy in the bond and in bond-breaking (exothermic reaction), all this energy is released back into the system. Give or take.

@Chestermiller gave an excellent quantitative approach. My explanation probably made some physical chemists squirm.

 

[Chestermiller]

Endothermic -> absorbs heat
Exothermic -> expels heat

The loss/gain of to the surroundings is then measured as a temperature decrease/increase. In bond-making (endothermic reaction), the majority of absorbed heat is stored as potential energy in the bond and in bond-breaking (exothermic reaction), all this energy is released back into the system. Give or take.

@Chestermiller gave an excellent quantitative approach. My explanation probably made some physical chemists squirm.

The first thing I would do would be to check to see which direction the reaction would proceed from the initial state.

 

[mjc123]

Energy is not "stored" in chemical bonds. Bonding is a low-energy state (by definition, or molecules would fall apart spontaneously). Bond breaking is endothermic (as stated in post #2) and bond making is exothermic.

 

[Mayhem]

Energy is not "stored" in chemical bonds. Bonding is a low-energy state (by definition, or molecules would fall apart spontaneously). Bond breaking is endothermic (as stated in post #2) and bond making is exothermic.

Why can't it be viewed as storing potential energy? Classically, it's analogous to saying "Walking up to the 12th floor does not 'store energy in you'. Being on the ground floor is a low-energy state (by definition, or I would start floating upwards spontaneously) ... "

I don't know. It's early.

 

[mjc123]

Are you comparing the bonded state to being on the 12th floor or the ground floor? That appears to be the confusion.