# Question on Thermodynamics from a recently conducted Examination

This is a question from the recently conducted JEE Main exam (in India). I had solved it myself, but found the answer to be incorrect. Here's the question.

Here is how I solved it:

$$\Delta H = \pu{41.1 kJ mol-1}$$
$$R = \pu{8.314 J mol-1 K-1}$$
$$w=\pu{36 g}$$ $$\Delta n_g=2$$ $$\ce{\Delta H = \Delta U + \Delta n_gRT}$$ $$\ce{ \Delta U = (44.1 - (2×8.31×373)/1000) \pu{J K-1}}$$ $$\ce{\Delta U = \pu{(41.1 - 6.199) J K-1}}$$ $$\ce{\Delta U = \pu{34.9 J K-1}}$$ $$Ans. 35$$

I feel that the answer can be 38 only if there is 1 mole of water instead of 2.

Kindly help, and point out my mistake, if any.

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Your mistake is that you took $$\Delta n_g=2$$ when it should be $$1$$. Remember it is asking for the internal energy change for the reaction (this can be seen by the fact that kJ/mol is being asked, not kJ-remember dividing by moles makes this intensive so there is no dependency on number of moles taken). Thus, the moles of water taken is irrelevant and $$\Delta n_g$$ only depends upon the stoichiometric coefficients. Since the reaction is $$\ce{H2O(l) -> H2O(g)}$$, so $$\Delta n_g$$=1.The mass of water given is just the examiner trying to trick(or rather, test) you.
2. Additionally, it should be made explicit that internal energy of vapourisation of the reaction $$\ce{H2O(l) -> H2O(g)}$$ is being asked because it can vary depending on if it is $$\ce{H2O(l) -> H2O(g)}$$ or $$\ce{2H2O(l) -> 2H2O(g)}$$ or $$\ce{3H2O(l) -> 3H2O(g)}$$ etc.(see addendum for more)
However, some texts define enthalpy of vapourisation and internal energy of vapourisation as the $$\Delta U$$ and $$\Delta H$$ of the reaction when 1 mol of the compound is undergoing phase transition, and then the above 2 points don't hold because if moles are taken, then stoichiometric coefficient magnitudes are irrelevant and only their ratios are important.
Just in case you want some depth in this topic, remember that in any calculation of $$\Delta _rH$$ and $$\Delta _r U$$ we assume the number of moles of the reactants/products to be same as that of the stoichiometric coefficients. This is because by definition, $$\Delta _rH$$ and $$\Delta _r U$$ depend only on the reaction equation and not on particular reactions in which we can vary the number of moles being taken.If you calculate $$\Delta H$$ or $$\Delta U$$ of a particular reaction, then that cannot be defined as $$\Delta _rH$$ or $$\Delta _r U$$ unless the number of moles of every reactant and product matches the stoichiometric coefficients. Finally, note that stoichiometric coefficients need not be the simplest ratios i.e. $$\Delta _rH$$ will be different for each of $$\ce{H2O(l) -> H2O(g)}$$, $$\ce{2H2O(l) -> 2H2O(g)}$$ and $$\ce{3H2O(l) -> 3H2O(g)}$$ because the stoichiometric coefficients are different.