Question

The enthalpy of vaporization of water at ${100^o}C$ is 40.63 KJ $mo{l^{ - 1}}$. The value $\Delta E$ for this process would be:
A. 37.53 KJ $mo{l^{ - 1}}$
B. 39.08 KJ $mo{l^{ - 1}}$
C. 42.19 KJ $mo{l^{ - 1}}$
D. 43.73 KJ $mo{l^{ - 1}}$

Answer 1

The enthalpy of vaporization is the amount of energy that must be added to the liquid substance in order to transform a quantity of that substance into gas. We can calculate $\Delta H$ from the measured value of $\Delta E$ with the ideal gas law, $PV = nRT$.

Complete step-by-step answer:
At constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. We can write it as $H = E + PV$.
As we know for an ideal gas, $PV = nRT$, we can further write it as $H = E + nRT$.
For vaporization of water, liquid water changes to vapour form. The chemical equation is as follows:
${H_2}{O_{(l)}} \rightleftharpoons {H_2}{O_{(g)}}$
For this process, change in enthalpy can be described as,
$\Delta H = \Delta E + \Delta {n_g}RT$
$\therefore \Delta E = \Delta H - \Delta {n_g}RT$
Where, $\Delta E$ is the change in internal energy of the system,
$\Delta H$ is the change in enthalpy of the system, R is the gas constant (8.314 J $mo{l^{ - 1}}$), temperature given ${100^o}C$ which is equal to 373K and $\Delta {n_g}$ is number of gaseous molecules of products – number of gaseous molecules of reactants.
Putting the values of enthalpy, gas constant, temperature and number of moles of gas in it, we get
$\Delta E = 40.63 - 1 \times 8.314 \times 373$
Thus, $\Delta E = 37.53$KJ $mo{l^{ - 1}}$

Hence, the correct option is (A).

Note: It is also the first law of thermodynamics. If the enthalpy has a negative value, it is an exothermic reaction but if it is positive, it is an endothermic reaction. Enthalpy of vaporization of water is positive, that means heat is absorbed in the reaction.