- #1
ElectroMaster88
- 5
- 3
- 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?
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?
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