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  • Standard enthalpy of vapourisation   Ho for water at 100o

Standard enthalpy of vapourisation \Delta_{\text{vap}}  Ho for water at 100oC is 40.66 kJ mol-1. The internal energy of vaporization of water at 100oC (in kJmol-1) is:

Option: 1

+ 37.56


Option: 2

- 43.76


Option: 3

+ 43.76


Option: 4

+ 40.66


Answers (1)

best_answer

\begin{array}{l} Given\ \Delta_{\mathrm{vap}} H^{\circ}=40.66 \mathrm{kJ} \mathrm{mol}^{-1} \\ T=100+273=373 \mathrm{K}, \Delta E=? \\ \Delta H=\Delta E+\Delta n_{g} R T \\\Rightarrow \Delta E=\Delta H-\Delta n_{g} R T \end{array}

\Delta \mathrm{n}_{\mathrm{g}}= number of gaseous moles of products - number of gaseous moles of reactants

\begin{array}{l} \mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{(g)} \\ \Delta n_{g}=1-0=1 \\ \Delta E=\Delta H-R T \\ \Delta E=\left(40.66 \times 10^{3}\right)-(8.314 \times 373) \\ =37559 \mathrm{J} / \mathrm{mol} \text { or } 37.56 \mathrm{kJ} / \mathrm{mol} \end{array}

 

 

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vishal kumar

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