Ncert Exercises & Solutions

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?

Given that, quantity of water

= 18.0 g

, pressure = 1 bar

As we know that,

18.0 g H_{2} O = 1 mole H_{2} O

Enthalpy change for vaporising 1 mole of

H_{2} O = 40.79 kJ {mol}^{- 1}

∴

Enthalpy change for vaporising 2 moles of

H_{2} O = 2 \times 40.79 kJ = 81.358 kJ

Standard enthalpy of vaporisation at

100 ° C

and 1 bar pressure,

∆_{vap} H ° = + 40.79 kJ {mol}^{- 1}

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