Thermodynamics Practice

Slick Rick is offering a “good deal” on platinum coins. He proffers a one ounce (28.3 g) platinum coin to you for a mere $175, a real steal. You know that he is smart enough not to try to slip you a different weight coin, but you suspect that perhaps the coin is just plated with platinum. You “accidentally” drop the coin in your precisely 50.0 mL of coffee in a Styrofoam cup that you only drink at 60.0C. If the temperature drops to 57.6C, is the coin likely real?

Specific Heat of Platinum: 0.13 J/gKSpecific Heat of Water: 4.184 J/gK

What is the specific heat of the coin? Is it the value of platinum?

Hint: This will not be given on the exam. It is the conversion factor between J and calories though.

Finding Specific HeatFirst Law of Thermo:

E = 0 = q water + q metal + q cal

0kJ (coffee cup)

q metal = -q water q metal = mmCmTm

q water = mwCwTw

mmCmTm= - mwCwTwCm= - mwCwTw mmTm

Cm= - 50.0g (4.184 g/JC) (60.0C - 57.6C) 28.3 g (57.6C - 21C)

= .48 g / J C

Lithium chloride dissolving in water is an exothermic reaction. If 3.45 g of solid lithium chloride is dissolved in

85.7 mL of water in a coffee cup calorimeter, and the temperature of the water goes from 21.0C to 29.6C,

what is the Hsol for LiCl?

LiCl(s) LiCl(aq)First Law of Thermo:

E = 0 = q water + q rxn + q cal

q rxn = -q water

q rxn = n Hrxn

q water = mwCwTw

0kJ (coffee cup)

Note: This looks exactly like other phase changes!

What amount of energy was absorbed by the water?

Trinitroglycerin, C3H5N3O9, is an explosive. The enthalpy of decomposition (below) at 1 atm pressure is -1541.4

kJ/mol. What is the Hf for nitroglycerin?C3H5N3O9(l) N2(g) + CO2(g) + H2O(l) + O2(g)

Hf Energy (kJ/mol)

CO2 (g) -393.5

H2O (g) -241.8

H2O (l) -285.8Need to balance the reaction!

C3H5N3O9(l) 1.5N2(g) + 3CO2(g) + 2.5H2O(l) + .25O2(g)

C3H5N3O9(l) 1.5N2(g) + 3CO2(g) + 2.5H2O(l) + .25O2(g)

Hrxn = (Hfprod) – (Hf react) = = 1.5 Hf N2 + 3Hf CO2 + 2.5Hf H2O + .25 Hf O2

– Hf Nitro

-1541.4 kJ =(0) + 3mol(-393.5 kJ/mol) + 2.5mol (-285.8kJ/mol) + 0 – Hf Nitro

Hf Nitro = -353.6 kJ/mol

The heat of combustion of fructose, C6H12O6, is -2812 kJ/mol. If a fresh golden delicious apple

weighing 4.23 oz (120 g) contains 16.0 g of fructose, what Caloric content does the fructose contribute to

the apple?

C6H12O6 + 6O2(g) 6CO2(g) + 6H2O(l)

Hf Energy (kJ/mol)

CO2 (g) -393.5

H2O (g) -241.8

H2O (l) -285.81 calorie = 4.184 J; 1 Dietary Calorie = 4.184 kJ

Hess’s Law

H2(g) + F2(g) 2HF(g)C(s) + 2F2(g) CF4(g)

2C(s) + 2H2(g) C2H4(g)

H = -537kJH = -680kJH = +52.3kJ

Using the following reactions and enthalpies:

What is the Hrxn for the following process (give answer in kJ):

C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g)