Question

It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (MW: 46.07 g/mol). What will be ΔSsys for the vaporization of 8.00 g of ethanol at 79.6 °C

Answer 1

Answer:

$\Delta S_{sys}=0.020kJ=20J$

Explanation:

Hello!

In this case, given the required energy to vaporize 1.00 mol of ethanol as the enthalpy of vaporization:

$\Delta H_{vap}=\frac{38.6kJ}{1.00mol}=38.6kJ/mol$

We can compute the entropy of the system for the vaporization of 8.00 g of ethanol, by first computing the moles:

$n_{et}=8.00g*\frac{1mol}{46.07g} =0.174mol$

And then setting up the following expression:

$\Delta S_{sys}=\frac{n_{et}*\Delta H_{vap}}{T}$

Whereas the temperature is in kelvins; thus, we obtain:

$\Delta S_{sys}=\frac{0.174mol*38.6\frac{kJ}{mol} }{79.6+273.15K}\\\\\Delta S_{sys}=0.020kJ=20J$

Best regards!