I have this reaction (all substances are gaseous):

$$ \begin{align} \ce{A + 4.5 B &→ C + 2 D + 2 E}\tag{R1}\\ \ce{C + 7.5 B &→ 8 D + 2 E}\tag{R2} \end{align} $$

Does the volume change when the reactions has occurred? Because if so, the concentration changes too, right?

  • 1
    $\begingroup$ The volume changes can only be predicted if at least one of the substances A, B, C, D and E is gaseous. If the substances are not gaseous, no predictions can be made. $\endgroup$
    – Maurice
    Nov 18, 2019 at 11:04
  • $\begingroup$ Just edited it(all substances are gaseous) $\endgroup$
    – Andrew
    Nov 18, 2019 at 11:11
  • $\begingroup$ I don't understand why sometimes questions that classify as homework (of course they might even be genuine users Q) get answered. OP poses a trivial question without attempts/discussion at all. $\endgroup$
    – Alchimista
    Nov 19, 2019 at 9:47
  • $\begingroup$ @Alchimista To poorly paraphrase Michael Caine, "...some men just want to watch the world learn". $\endgroup$
    – J. Ari
    Nov 19, 2019 at 19:47

1 Answer 1


A volume change from a reaction occurs if the difference in mole of gases does not equal to $0$

In order to find the answer for your question, we will have to add both reactions together, resulting in: $$\ce{A + 4.5 B + C + 7.5 B -> C + 2 D + 2 E + 8 D + 2E}$$ We can shorten the reaction by adding up all duplicates and removing both $\ce{C}$ from each side, resulting in: $$\ce{A + 12 B -> 10 D + 4E}$$ Now if we calculate the difference in mole gasses we get: $$\Delta n = \pu{(10 + 4)mol} - \pu{(1 + 12)mol} = \pu{1 mol}$$ Based on the positiv difference in mole of gasses we can say that the reaction will change in volume, that is it will expand.


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