The standard enthalpy change of vaporization of water is + 40.7 kJ mol^-1. Calculate the enthalpy change (in kJ) when 15 g of water vaporizes at its boiling point at 1 bar.

Question

Quizplus · Accepted Answer

First, we need to convert 15 g of water to moles. 
Molar mass of water (H2O) = 18.015 g/mol
Number of moles (n) = mass/molar mass = 15 g/18.015 g/mol = 0.8329 mol

Now, we can use the formula:
Enthalpy change (ΔH) = n × ΔHvap

Substitute the given values:
ΔH = 0.8329 mol × 40.7 kJ mol⁻¹ = 33.9 kJ

Since the question asks for the answer in kJ, we round off to one decimal place:
ΔH = +34 kJ

Therefore, the correct choice is C.

The Standard Enthalpy Change of Vaporization of Water Is