Question

What is the entropy change of the system when 17.5 g of liquid benzene (C₆H₆) evaporates at the normal boiling point? The normal boiling point of benzene is 80.1�C and ?H _vap is 30.7 kJ/mol.

	�19.5 J/K
	+85.9 J/K
	25.2 J/K
	+19.5 J/K

Answer 1

Expression for change in entropy (ΔS) in terms of heat(q in joule) and temperature(T in kelvin) can be written as follows:

ΔS = q / T

Temperature in Kelvin can be calculated as follows:

T = 273 + 80.1 K = 353.1K

Heat in terms of joules can be calculated as follows:

q = (17.5 g) (1 mol benzene / 78.11g of benzene) (30.7 KJ / 1 mol)(1000J / 1KJ) = 6878 J

Thus, Change in entropy can be calculated as follows:

ΔS = 6878 J / 353.1K = 19.5 J/K

Thus, change in entropy of the given system is +19.5 J/K