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 correct option is A $37.53$ KJ $mo{l}^{-1}$

Solution:- (A) $37.53KJ/mol$

We know that for gaseous reactants and products , we have a relation between standard enthalpy of vapourization $\left({\Delta H}_{vap.}\right)$ and standard internal energy $\left(\Delta E\right)$ as-

${\Delta H}_{vap.}=\Delta E+\Delta n_{g}RT$

whereas,

$\Delta n_g=n_2-n_1$, i.e., difference between no. of moles of reactant and product.

For vapourization of water,

${H_{2}O}_{\left(l\right)}⟶{H_{2}O}_{\left(g\right)}$

$\therefore \Delta n_g=1-0=1$

$T=100℃=(100+273)=373K$

$\therefore {\Delta H}_{vap.}=\Delta E+\Delta n_{g}RT$

$\Rightarrow \Delta E={\Delta H}_{vap.}-\Delta n_{g}RT=40.63-(1\times 8.314\times {10}^{-3}\times 373)=37.53KJ/mol$

Hence the value $\Delta E$ for this process will be $37.53KJ/mol$.