Calculate the heat energy released when 22.6 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point.

Question

Constants for mercury at 1 atm

heat capacity of Hg(l) 28.0 J/(mol⋅K)

melting point 234.32 K

enthalpy of fusion2.29 kJ/mol

Answer 1

These problems are best answered using a step-wise approach.

q = mC∆T (no phase change)

q = m∆H (phase change)

Step 1: lower temp of liquid from 25º (298K) to 234.32K

q = mC∆T = (22.6 g/200.59g/mol)(28.0 J/molK)(64K) = 201.9 J = 0.202 kJ

Step 2: change 22.6 g (0.113 moles) Hg from liquid to solid at 234K (phase change)

q = m∆Hfusion = (0.113 mol)(2.29 kJ/mol = 0.2580 kJ

Add up the heat from both steps:

0.202 kJ + 0.258 kJ

Heat released = 0.460 kJ (it would have a negative sign but since we state that it is "released" that is given)