What direction would equilibrium moves towards based on the following if we increased the volume of the container.

[tex]2A_{(g)} + 5B_{(g)} + 12C_{(g)}[/tex] ↔ [tex]14AC_{(g)} + 5B_{(s)}[/tex]

Answer choices:
a) reactants
b) no change
c) products
d) decrease in volume

Please help!

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TheSection TheSection · Answer 1 · 2021-07-21T20:10:59+02:00

To answer this question, we will first find out the number of gaseous moles on each side of the equilibrium

on the left:

we have 2 moles of A, 5 moles of B and 12 moles of C

which gives us a grand total of 19 gaseous moles

on the right:

here, we have 14 moles of AC gas, we will not count the number of moles of B because it's a solid

giving us 14 gaseous moles on the right

Where does the reaction shift?

more gaseous moles means more space taken, because gas likes to fill all the space it can

if we have more volume, more gas can move around without colliding (reacting) with each other

Hence more volume favors the side with more gaseous moles

here, the left has more gaseous moles. So we can say that the reaction will shift towards the left, or the reactants side

Asdfjkadfjkakjf Asdfjkadfjkakjf · Answer 2 · 2021-07-22T10:56:03+02:00

Answer:

Explanation:

given reversible chemical reaction:

2A(g) + 5B(g) + 12C(g) ↔ 14AC(g) + 5B(s)

chemicals in solid form do not take up a lot of volume so change in container volume has no effect

look at chemicals in gas form only:

the total no. of moles of reactants in gas form = 2 + 5 + 12 = 19

the total no. of moles of products in gas form = 14

so an increase in volume of the container will favor the reaction direction with higher volume n high volume means higher no. of moles

the ans is the equilibrium will move towards a) reactants