Since the dH reaction is +312 kJ, that means the reaction is endothermic so I would write it as
C2H6 + heat = C2H2 + 2H2
So if T is decreased, rxn shift to the left. (Rxn uses heat to go to the right; take heat away it moves to the left).
Add He at constant volume. If volume does not change, there is no change in concentration (still same n and L so mols/L = constant) and there is no shift.
If He added at constant pressure, that means volume must increase to keep P from rising. If volume increases, reaction shifts to larger number of molecules (to the right).
C2H6(g) <=> C2H2(g) + 2 H2(g)
The ΔH° for the reaction above is +312 kJ. The system is initially at equilibrium. In which direction will the reaction shift in each of the following situations?
e) The temperature is decreased.
g) He is added at constant volume.
h) He is added at constant pressure.
(Possible Answers: forward, reverse, no shift)
My Work:
I did a-d and f, these are the ones that I have no clue how to do.
I do not understand how the +312 kJ affects anything. I thought decreasing temperature wouldn't cause a shift, but the assignment says I'm wrong. Why does it shift one way or the other?
Also, how can helium just be added to the reaction? And how does that effect the rest of the compounds involved?
1 answer