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 vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?

Question

Philoid · Accepted Answer

o The enthalpy or ∆H of a reaction is defined as the change in energy (more specifically heat energy) for one mole of a substance involved in a process.

It is given that, the enthalpy change for 18g of water (at 100^⁰ and 1 bar pressure) is =40.79 kJ mol^–1 in the process of vapourisation.

And for water, 18g is =1 mole (For H₂O, the calculation of molar mass: 2×1 + 16 = 18g)

Hence, enthalpy change of vapourisation for 1 mole = 40.79 kJ mol^–1 enthalpy change of vapourisation for 2 moles of water = (40.79 × 2) = 81.58kJ mol^–1

o Standard enthalpy of a process or ∆H⁰ is the enthalpy change of the process when involved substances are in their standard states.

Hence, for water ∆H_{vapourisation,} ⁰ will be equal to =40.79 kJ mol^–1

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 vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?

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 vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?