Calculate q, w, ΔE, and ΔH, when 175 g of water evaporates at 1.00 atm and 100.0 °C. The heat of vaporization of water is 40.7 kJ/mol. q= kJ w= kJ ΔE= kJ ΔH= kJ

q= 395.6 kJ

w= 30.1429 kJ

ΔE= 365.4571 kJ

ΔH= 395.6 kJ

Explanation:

First law of thermodynamics:

ΔE = q - w

(q is taken positive when enters to the system and w is taken positive when it is applied to the surroundings)

number of moles = mass / molar mass = 175/18 = 9.72 mol of water

ΔH = number of moles * heat of vaporization = 9.72*40.7 = 395.6 kJ

At constant pressure:

q = ΔH = 395.6 kJ

When water evaporates, it is expanded, then the work is made on the surroundings. At constant pressure:

W = n*R*T

where n is the number of moles, R is the gas constant (8.314 J/(mol*K)) and T is the temperature (in K)

W = 9.72*8.314*(273 + 100) = 30142.9 J = 30.1429 kJ

Finally:

ΔE = q - w = 395.6 - 30.1429 = 365.4571 kJ

i think that the answer is c