For the vaporization of water at 1.00 atm,
ΔH = 43.54 kJ/mol at 298 K and ΔH = 40.68 kJ/mol at 373 K
The constant-pressure heat capacity of liquid water is 75.3 J/mol • K. Calculate the constant-pressure heat capacity for H₂O(g) .

Question

Quizplus · Accepted Answer

First, we can use the formula ΔH = q + w, where q is the heat transferred and w is the work done. At constant pressure (which is the case for water vaporization at 1 atm) w = -PΔV = -ΔnRT = -nΔH, where n is the number of moles of water vaporized, R is the gas constant, and T is the temperature in Kelvin. Therefore, ΔH = q - nΔH, or q = 2ΔH. Next, we can use the formula q = nCΔT, where C is the heat capacity and ΔT is the change in temperature. Since the vaporization occurs at constant temperature, we have q = nCΔT = nC(373-298) = nC(75). Setting these two expressions for q equal to each other and solving for C gives: 2ΔH = nCΔT C = 2ΔH / (nΔT) = 2ΔH / R At 298 K: C = 2(43.54 kJ/mol) / (8.314 J/mol•K) = 104.7 J/mol•K At 373 K: C = 2(40.68 kJ/mol) / (8.314 J/mol•K) = 97.8 J/mol•K Taking the average of these values gives C = 101.25 J/mol•K. However, we must remember that we are looking for the heat capacity of H₂O(g), not water. The heat capacities of the two substances can be assumed to be comparable, since they have similar molecular weights and are both composed of H, S, U, and O atoms. Therefore, the answer is D) 37.2 J/mol•K, which is closest to the heat capacity of liquid water (75.3 J/mol•K) but lower due to the decreased number of degrees of freedom in the gas phase.

For the Vaporization of Water at 1.00 Atm