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Homework 5
1.) Given the equation: 2Ag2S(s) + 2H2O(l) → 4 Ag(s) + 2H2S (g) + O2 (g) H = +595.5 kJ
Calculate the H for the following reaction:
H =
2Ag(s) + H2S (g) + ½ O2 (g) → Ag2S(s) + H2O (l)
The original equation is reversed: 4 Ag(s) + 2H2S (g) + O2 (g) → 2Ag2S(s) + 2H2O(l)
And then multiplied by ½
H = -595.5 kJ
2Ag(s) + H2S (g) + ½ O2 (g) → Ag2S (s) + H2O (l) H = (-595.5/2) kJ
H = - 297.8 kJ
2.) Consider the following reaction:
H = -127 kJ
Ag (s) + ½ Cl2 (g) → AgCl (s)
a.) Is this reaction exo or endothermic – how do you know?
The reaction is exothermic as
indicated by the – sign on the ΔH value
b.) What quantity of heat is needed to decompose 102.5 grams of solid silver chloride INTO its
elements?
102.5 grams AgCl x
1 mole AgCl
127 kJ
=
x
143.4 grams AgCl 1 mole AgCl
90.8 kJ of heat is needed
3.) When 25.0 mL of 0.500 M H2SO4 is added to 25.0 mL of dilute M KOH (both initially at 23.5°C) in a
coffee-cup calorimeter, the temperature rises to 30.2°C. You may assume that the heat capacity for
the solution = 4.814 J/goC
From this data:
a. Find qrxn for the process. (again, because the solutions are dilute, you can assume the solution
has a density of 1.00 g/mL)
b. Find ∆Hrxn, in kJ/mol H2SO4 for the reaction (be sure to get the correct sign of ∆H).
You can assume the mass of the solution ~ mass of water due to the solutions being so dilute
~25.0 mL of water from H2SO4 + ~25.0 mL of water
from KOH
= essentially 50.0 mL of water!
50.0 mL x 1.00 grams/mL = 50.0 grams of water
qwater = mwaterCwaterTwater
qwater = 50.0 grams (4.184 J/goC)(30.2oC – 23.5oC)
qwater = 50.0 grams (4.184 J/goC)(6.7oC)
qwater = 1.40 x 103 J
Therefore, the qrxn is equal to the qwater but of opposite sign – therefore qrxn
= -1.40 x 103 J
To calculate the kJ/mol we will simply determine the kJ from the q rxn and then determine the moles of
H2SO4 present – and then literally divide the kJ by the moles!
 1.40 x 10 3 J x
1 kJ
 1.40 kJ
1000 J
- 1.40 kJ
 112 kJ/mole
0.0125 moles H 2 SO 4
0.500 mole H 2SO 4
1L
x
x 25.0 mL  0.0125 moles H 2SO 4
1L
1000 mL
-112 kJ/mole
4.) Pioneers used to reheat their coffee by placing and iron poker from the fire directly into their cup. If
the cup initially held 0.500 L of coffee at 20.1°C, what would be the final temperature of the coffee
when a 5.00 x 102 g iron poker at 402.3 °C was placed in it? Assume that no heat is lost to the
surroundings during the process, and that the heat capacity and density of coffee is that same as
that of water (I’m sure it wasn’t very good coffee!). (note: you will need to look up the heat
capacities in your text for the species in this problem!)
0.500 L x
1000 mL 1.00 g
x
= 5.00 x 102 grams coffee ≈ water
1L
1.00 mL
-qlost = q gained
-mCT = mCT
and 5.00 x 102 grams Fe
-mFeCFeTFe = +mwaterCwaterTwater
We put the – sign on the Fe because it will be losing heat as it cools and the + side is the water
because it will be gaining the heat
-[5.00 x 102 grams x (0.451 J/goC)(Tfinal - Tinital)] = +[5.00 x 102 grams x (4.184 J/goC)(Tfinal - Tinital)]
- [5.00 x 102 grams x (0.451 J/goC)(Tfinal – 402.3o)] = +[5.00 x 102 grams x (4.184 J/goC)(Tfinal – 20.1o)]
Condense the math a bit; and then call Tfinal = X for simplification
-225.5 J/oC (Tfinal – 402.3oC) = 2092 J/oC (Tfinal – 20.1oC)
Use the distributive property:
-225.5 J/oC (X) + 90718.65 J = 2092 J/oC (X) – 42049.2 J
Condense like terms:
90718.65 J + 42049.2 J = 2092 J/oC (X) + 225.5 J/oC (X)
132767.85 J = 2317.5 J/oC (X)
Solve for X : which recall the variable for Tfinal
= 57.3oC
5.) Determine the following:
a. The wavelength of light (in nanometers) if the frequency is 8.68 x 10-8 sec-1
b. The frequency of light if the wavelength is 901 nm
=
3.45 x 1015 nm
=
=
2.20 x 10-19 J
3.33 x 1014 Hz