Le Chatliers Principle Destroyer

[Molarcule]

I need some clarification here

#68Which of the following would shift the equilibrium to the right?
2 NOBr (g)--> 2 NO (g) + Br2(g) Detla(H)=30KJ

A)Increase Pressure
B)Decrease temperature
C)Remove some NOBr
D)Increase the container volume
E)Two of the above

Here is my logic

A)If we increase pressure there would be more collisions hence more products will form (Shifting right)
B)If we decrease the temperature, to compensate (Shift Left)
C)We would shift towards the left to compensate for the missing NOBr. (Shift left)
D)If we increase the container volume there will be less collisions having more reactants (Shift left)

The answer is D. Why am I wrong?

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[Baylor2011]

A)If we increase pressure there would be more collisions hence more products will form (Shifting right)
B)If we decrease the temperature, to compensate (Shift Left)
C)We would shift towards the left to compensate for the missing NOBr. (Shift left)
D)If we increase the container volume there will be less collisions having more reactants (Shift left)

If you increase container volume, it's going to favor the side of the reaction that has more moles of gas - therefore filling up more of the container. Conversely, if you decreased the container's volume, the reaction would favor the side of the reaction which has the least moles of gas.

Not A, since if you increase pressure it's not going to want to shift to the side of the reaction that has more moles of gas, it's going to favor the side that has the least moles of gas

You have B and C correct - B because since it's endothermic (positive enthalpy) you can in effect write "heat" in as a reactant, so if you decrease temperature, it's going to favor the side of the reaction that has heat. C you're right on the spot.

 

[nze82]

I need some clarification here

#68Which of the following would shift the equilibrium to the right?
2 NOBr (g)--> 2 NO (g) + Br2(g) Detla(H)=30KJ

A)Increase Pressure>>>The rxn shifts to the side with fewer number of gas moles (Left)
B)Decrease temperature>>>The rxn is Endothermic, so heat is considered a reactant. Remove that reactant and the rxn will move to the side that will compensate for this change (Left)
C)Remove some NOBr>>>Same logic as above.
D)Increase the container volume>>>Remember PV = nRT?! According to this formula if you increase the volume the pressure drops to counteract the change in volume (because PV must remain constant).To compensate for the drop in pressure the rxn moves to the side with greater number of gas moles (RIGHT).
E)Two of the above

Here is my logic

A)If we increase pressure there would be more collisions hence more products will form (Shifting right)
B)If we decrease the temperature, to compensate (Shift Left)
C)We would shift towards the left to compensate for the missing NOBr. (Shift left)
D)If we increase the container volume there will be less collisions having more reactants (Shift left)

The answer is D. Why am I wrong?

Good luck