Calculate the heat energy released when 14.0 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) is 28.0 J/mol*K

melting point is 234.32 K

enthalpy of fusion is 2.29 kJ/mol

Answer 1

1.) Find how many moles of Mercury we have:

14.0 g Hg x 1 mol Hg/201 g = 0.0697 moles

2.) Determine the heat released as it goes to a liquid: Note: 25ºC = 298.15K

q = heat = m x C x ∆T = (0.0697 mol)(28.0 J/molK)(63.83K) = 124.6 J

Note:

25ºC = 298.15K

∆T = 234.32 K- 298.15 K = 63.83 K

3.) Determine the heat released for 14.0 g Hg(l) at 234.32K turning to Hg(s)

q = m x ∆H fusion = (0.0697 mol)(2.29 kJ/mol) = 0.1596 kJ = 159.6 J

4.) Calculate the total Energy:

124.6 J + 159.6 J = 284.2 J of heat released

Answer 2

14.0 g Hg x 1 mol Hg/201 g = 0.0697 moles

Steps:

(1) heat released for 14.0 g Hg(l) at 25º going to Hg(l) at 234.32K (25ºC = 298.15K)

(2) heat released for 14.0 g Hg(l) at 234.32K turning to Hg(s)

(3) sum the heats released

(1) q = heat = mC∆T = (0.0697 mol)(28.0 J/molK)(63.83K) = 124.6 J

(2) q = m x ∆Hfusion = (0.0697 mol)(2.29 kJ/mol) = 0.1596 kJ = 159.6 J

(3) Sum = 124.6 J + 159.6 J = 284.2 J of heat released