High School

It takes 38.6 kJ of heat at constant pressure to vaporize 1.00 mol of ethanol (MW: 46.07 g/mol). What will be [tex]\Delta S_{sys}[/tex] (in J/K) for the vaporization of 6.50 g of ethanol at 78.0°C?

Answer is in J/K.

Answer :

To find the change in entropy for the vaporization of 6.50 g of ethanol at 78.0°C, calculate the number of moles, use the given heat of vaporization, then divide the total heat for 6.50 g by the temperature in Kelvin. The result is 15.5 J/K.

To find the change in entropy (">ΔS<") for the vaporization of 6.50 g of ethanol at 78.0°C, we need to use the given heat required to vaporize 1.00 mol of ethanol and the molecular weight (MW) of ethanol.

The given heat of vaporization is 38.6 kJ for 1 mol, which is equivalent to 38600 J.

Firstly, calculate the number of moles of ethanol in 6.50 g using the molecular weight:

Moles of ethanol = 6.50 g / 46.07 g/mol = 0.141 mol

Then, calculate the total heat absorbed for vaporizing 6.50 g of ethanol:

Heat absorbed (q) = 0.141 mol * 38600 J/mol = 5442.6 J

The change in entropy under constant pressure can then be calculated using the formula:

ΔS = q / T

Given temperature T is 78.0°C or 351 K (converting to Kelvin by adding 273 to the Celsius temperature).

So,

ΔS = 5442.6 J / 351 K

= 15.5 J/K

The change in entropy (">ΔS<") for the vaporization of 6.50 g of ethanol at 78.0°C is therefore 15.5 J/K.