Name: _________________________________________
Date:____________________________
Optional Homework 2 – for practice only!!
1.) State the first law of thermodynamics in your OWN words
Energy can neither be created nor destroyed, it just changes form. Energy cyles, changing from PE to KE
over and over and over again.
2.) Find q, w, and ∆E for the following systems:
a. A substance gains 9.6 J of heat and does 15.3 J of work ∆E = 9.6 – 15.3 = - 5.7 J
q = + 9.6 J w = -15.3 J
b. A gas loses 425.3 J of heat while doing 112.0 J of work ∆E = -425.3 – 112.0 = - 537.3 J
q = -425.3 J
c.
w = -112.0 J
A substance has 350.0 J of work done on it and absorbing 82.9 J of heat ∆E = 82.9 + 350.0 = 432.9 J
q = +82.9 J
w = +350.0 J
d. A gas has 110.0 J of work done on it while giving off 287.5 J of heat ∆E = -287.5 + 110.0 = -177.5 J
q = -287.5 J
w = +110.0 J
3.) For each of the following values of q and w, state whether the system (1) gained or lost heat, (2) did work or
had work done on it, (3) gained or lost energy
a.
q = +4.9 J, w = +7.8 J: gained heat, had work done on it, gained energy
b. q = -55.8 J, w = -17.3 J: lost heat, did work, lost energy
c.
q = + 254 J, w = -13 J: gained heat, did work, gained energy overall
d. q = -12.8 J, w = +6.4 J : lost heat, had word done on it, lost energy overall
4.) 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:
Ag(s) + ½ H2S (g) + ¼ O2 (g)
→ ½ Ag2S (s) + ½ H2O (l)
The original equation is reversed: 4 Ag(s) + 2H2S (g) + O2 (g) → 2Ag2S(s) + 2H2O(l)
And then multiplied by ¼
Ag(s) + ½ H2S (g) + ¼ O2 (g)
∆H = -595.5 kJ
→ ½ Ag2S (s) + ½ H2O (l) ∆H = -148.9 kJ
∆H = - 148.9 kJ
5.) How many moles of CH4 (g) release 1.00 x 106 kJ of heat in the following reaction?
CH4 (g) + O2 (g) → CO2 (g) + H2O(l) ∆H = -890.3 kJ
1.55 x 103 kJ x
1 mole CH 4
− 890.3 kJ
= 1.74 moles of CH4 Remember, even though the ∆H value is negative, in
this instance we must use the absolute value (only the numerical value) of
∆H otherwise we would get a negative number of moles and negative moles
(or grams etc. . . ) does NOT make sense!
6.) What volume of CH4 (g) does that correspond to at 25oC and 745 torr?
PV = nRT
V=
nRT
=
P
Latm
) (32 + 273.15 K)
mol K
=
1atm
755torr x
760 torr
1.74 moles (8.206x10 − 2
7.) Give formulas for the following molecular compounds:
a.
dinitrogen tetraoxide N2O4
b. sulfur hexachloride SCl6
c.
diiodine heptoxide I2O7
43.8 L
8.) Complete the following table:
IONS
nitrate
sulfite
hydrogen carbonate
phosphate
sodium
NaNO3
Na2SO3
NaHCO3
Na3PO4
ammonium
NH4NO3
(NH4)2SO3
NH4HCO3
(NH4)3PO4
calcium
Ca(NO3)2
CaSO3
Ca(HCO3)2
Ca3(PO4)2
aluminum
Al(NO3)3
Al2(SO3)3
Al(HCO3)3
AlPO4
5 And 6 Name the following compounds.
Species
Name
Species
Name
SrCl2
Strontium chloride
NO
Nitrogen monoxide
Ga2S3
Gallium sulfide
NO2
Nitrogen dioxide
Zn(ClO4)2
Zinc perchlorate
PBr5
Phosphorus pentabromide
CrO3
Chromium (VI) oxide
PBr3
Phosphorus tribromide