Question

The standard enthalpy of vapourization of water is 40.7 the change in internal energy and entropy

Answer 1

Answer:

6.48 kJ

Explanation:

The molar heat of vaporization, ΔHvap, sometimes called the molar enthalpy of vaporization, tells you how much energy is needed in order to boil 1 mole of a given substance at its boiling point.  

In water's case, a molar heat of vaporization of 40.66 kJ mol−1 means that you need to supply 40.66 kJ of heat in order to boil 1 mole of water at its normal boiling point, i.e. at 100∘C.  

ΔHvap=40.66 kJ.mol−1

You need 40.66 kJ of heat to boil 1 mole of water at its normal boiling point.

Now, the first thing to do here is to convert the mass of water to moles by using its molar mass

2.87g⋅

1 mole H2O

18.015g

=0.1593 moles H2O

You can now use the molar heat of vaporization as a conversion factor to determine how much heat would be needed to boil 0.1593 moles of water at its boiling point

0.1593moles H2O⋅

40.66 kJ

1mole H2O

=

6.48 kJ−−−−−−

The answer is rounded to three sig figs, the number of sig figs you have for the mass of the sample.

Explanation:

Answer 2

Explanation:

Molar enthalpy of vaporization

=2.275x18=40.95kj/mol

ΔvapS=ΔvapH/T

= 40.95x10^3/373.

= 109.78j/k/mol