Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • 18.0 g of water completely vapourises 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 vaporizing two moles of water under the same conditions?

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 vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?

Answers (1)

1 mol of water or 18g of water has an enthalpy of change for vaporisation. Now it is given that the change of enthalpy required for 1 mole of water is 40.79 kJ/mol.

For 2 moles of water, the enthalpy change will be 2 x 40.79 = 81.58 KJ . Therefore, the standard enthalpy of vaporisation of water is \Delta _{vap} H = 40.79\; kJ / mol

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question