18.0 g of water completely vaporises at 100°C and 1 bar pressure and the enthalpy change in the process is 40. 79 "kJ mol"^(-1). What will be the enthalpy change for vapourising two moles of water under the same conditions ? What is the standard enthalpy of vapourisation for water ?

18.0 g of water completely vaporises at 100°C and 1 bar pressure and

1.	18.0 g of water completely vaporises at 100°C and 1 bar pressure and the enthalpy change in the process is 40. 79 "kJ mol"^(-1). What will be the enthalpy change for vapourising two moles of water under the same conditions ? What is the standard enthalpy of vapourisation for water ?
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Quantity of water = 18.0 g, pressure <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> bar <br/> As we know that, `18.0 g H_(2) O = 1 "<a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a>" H_(2) O` <br/> <a href="https://interviewquestions.tuteehub.com/tag/enthalpy-15226" style="font-weight:bold;" target="_blank" title="Click to know more about ENTHALPY">ENTHALPY</a> change for vapourising l mole of `H_(2) O = 40.79 "kJ mol"^(-1)` <br/> `therefore` Enthalpy change for vapourising 2 moles of `H_(2) O = 2 xx 40.79 "kJ" = 81.358 "kJ"` <br/> Standard enthalpy of vaporisation at 100°C and 1 bar pressure, `Delta_("vap") H^(@) = + 40.79 "kJ mol"^(-1)`</body></html>