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.[![enter image description here](https://i.stack.imgur.com/XY532.jpg)](https://i.stack.imgur.com/XY532.jpg)


Here is how I solved it:


$\Delta H = \pu{41.1 kJ mol-1}$  

$R = \pu{8.314 J mol-1 K-1}$  

$w=\pu{36 g}$
$$ \Delta n\_g=2$$
$$ \ce{\Delta H = \Delta U + \Delta n\_gRT}$$
$$ \ce{ \Delta U = (44.1 - (2×8.31×373)/1000) \pu{J K-1}}$$
$$\ce{\Delta U = \pu{(41.1 - 6.199) J K-1}}$$
$$\ce{\Delta U = \pu{34.9 J K-1}}$$
$$Ans. 35$$


However, the answer key states that the answer will be 38, instead of 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.




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.


That being said, I might add that the question is poorly framed and also somewhat ambiguous. This is because of 2 reasons:-


1. It should be clearly mentioned that internal energy of vapourisation *of the reaction* is being asked and the only hint of this is kJ/mol which is easy to miss.
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.


Addendum:


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.


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