The enthalpy of vaporization of water at 100 | vedclass

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Question

(A) The chemical equation for the vaporization of water is: $H_2O_{(l)} 	o H_2O_{(g)}$The change in the number of moles of gas is $\Delta n_g = 1 - 0

Vedclass · Accepted Answer

$37.53$

Vedclass · Answer

(A) The chemical equation for the vaporization of water is: $H_2O_{(l)} 	o H_2O_{(g)}$The change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.The relationship between enthalpy change and internal energy change is given by: $\Delta H = \Delta U + \Delta n_g RT$.Given: $\Delta H = 40.63 \, kJ \, mol^{-1}$,$R = 8.314 	imes 10^{-3} \, kJ \, K^{-1} \, mol^{-1}$,and $T = 100 + 273 = 373 \, K$.Substituting the values: $40.63 = \Delta U + (1 	imes 8.314 	imes 10^{-3} 	imes 373)$.$40.63 = \Delta U + 3.10$.$\Delta U = 40.63 - 3.10 = 37.53 \, kJ \, mol^{-1}$.

The enthalpy of vaporization of water at $100 \, ^\circ C$ is $40.63 \, kJ \, mol^{-1}$. The value of $\Delta U$ for this process is ....... $kJ \, mol^{-1}$.

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