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.
