1 mole of each $\ce{CaC2}$, $\ce{Mg2C3}$, $\ce{Al4C3}$ react with excess water in separate open flask. Work done during dissolution shows the order

  1. $\ce{CaC2}=\ce{Mg2C3}<\ce{Al4C3}$
  2. $\ce{CaC2}=\ce{Mg2C3}=\ce{Al4C3}$
  3. $\ce{Mg2C3}<\ce{CaC2}<\ce{Al4C3}$
  4. $\ce{Mg2C3}<\ce{Al4C3}<\ce{CaC2}$

Calcium carbide gives ethyne on reaction with water. One mole of calcium carbide reacts with $2$ moles of water.

Magnesium carbide gives prop-1-yne on reaction with water. One mole of magnesium carbide reacts with $4$ moles of water.

Aluminum carbide gives methane on reaction with water and one mole of aluminium carbide reacts with $12$ moles of water.

Work done $=P\Delta V$. But since the amount of water is large compared to the amount of carbide, will the work done differ?

(Answer given is $1$)


The relevant equations are:

$\ce{Mg2C3(s) + 4 H2O (aq) -> C3H4 (g) + 2Mg(OH)2 (aq)}$

$\ce{Al4C3 (s) + 12H2O(aq) -> 3CH4(g) + 4Al(OH)3(aq) }$

$\ce{CaC2(s) + 2H2O(aq) -> C2H2(g) + Ca(OH)2(aq)}$

As you can see, most number of gaseous moles are released in the 2nd reaction while equal number of gaseous moles are released in the first and third reaction, therefore maximum work is done in the first reaction under same conditions of temperature and pressure.

Note that:

In open vessel reaction, gases expand against a constant external atmospheric pressure, thereby doing work on the surroundings , thus the work done $= P_{\text{atm}}\Delta V$ is independent of the gas's pressure and only dependent on $\Delta V$

Also, recall that Avogadro's law states:

For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.

Hence, here, $\Delta V$ only depends on the moles of the gas.

| improve this answer | |
  • 1
    $\begingroup$ Good answer! Though I think it's worth pointing out the question is slightly ambiguous. "Work done" by whom? Surroundings or system? Because the signs will be opposite for each, and the order will reverse itself. Fortunately though, none of the options for this question are like that, so we're safe. $\endgroup$ – Gaurang Tandon Mar 30 '18 at 1:54
  • $\begingroup$ @GaurangTandon we generally talk wrt system in chemistry... $\endgroup$ – Archer Mar 30 '18 at 1:59
  • $\begingroup$ Of course, we can assume that. $\endgroup$ – Gaurang Tandon Mar 30 '18 at 2:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.