4. Which one of the following is not a redox reaction? a. Na 6 FeCl 8 (s)+2Na(l)
ID: 894700 • Letter: 4
Question
4. Which one of the following is not a redox reaction?
a. Na6FeCl8(s)+2Na(l) -- > 8NaCl(s)+Fe(s)
b. Al(OH)3 (aq)+3H+ -- > Al3+(aq)+3H2O(l)
c. CuCl2(aq)+Ni(s) -- > Cu(s)+NiCl2(aq)
d. C6H12O6(s)+6O2(g) -- > 6CO2+6H2O(l)
e. CO2(g)+H2(g) -- > CO(g)+ H2O(g)
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5. A voltaic cell consists of a Zn/ZN2+ electrode and a Fe/Fe2+ electrode. If [ZN2+] = 0.005M and if [Fe2+] = .0500 M , what is Ecell?
a. .76 b. 0.25 c. 0.45 d. 0.31 e. 0.37
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6.2. The formation constant for the reaction Ag+ (aq)+2NH3(aq) < --- > Ag(NH3)2+(aq) is K=1.7x107 at 25 deg C. What is the G(kJ/mol) at this temperature?
Delta H delta G S
Ag+(aq) = 105.6 77.1 72.7
NH3(aq)= -80.3 -26.6 111
Ag(s) = 0 0 42.6
a. -23 b. -18 c. -3.5 d. -1.5 e. -41
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7.3. Nitric oxide reacts with chlorine to form NOCL. What is the free energy (kJ) of the reaction at 277 deg C? 2NO(g) + Cl2(g) ---> 2NOCl(g)
delta H delta G S
NO = 90.2 86.6 210.7
Cl2 = 0 0 223.0
NOCL=
a. +144 b.. -41.0 c. -10.3 d. -22.2 e. +41.0
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8.5. Estimate the boiling point (deg C) of phosphoric Acid.
H3PO4(s) < --- > H3PO4(l)
Hf kJ/mol -1279 - -127.17
Gf kJ/mol -1119 - 1123.6
Sf J/(k mol) 110.5 - 150.8
a. 42 b. 181 c. 315 d. -92 e. 305
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9.7. Calculate the equilibrium constant, K, for the following reaction at 25C:
CH4(g)+2H2O(g) < --- > CO2(g)+4H2(g) G= -113.6 kJ (please show ICE chart and explain process)
Delta H delta G S
H20(g) -241 -228.6 188.7
CO2 (g) -393.5 -394.4 213.6
H2 (g) 0 0 130.6
a. 8.2x1019 b. 0.58 c. 0.96 d. 1.4x10-46 e. 1.2x10-20
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10.8. Which of the following results in a decrease in the entropy of the system?
a. O2(g), 300K -- > O2(g), 400K b. H2O(s) -- >H2O(l)
c. N2(g) -- >N2(aq) d. NH3(l) -- >NH3(g)
e. 2H2O2(g) -- >2H2O(g)+O2(g)
Explanation / Answer
4) the reaction is not a redox reaction if oxidation and reduction does not takes place.
b. Al(OH)3 (aq)+3H+ -- > Al3+(aq)+3H2O(l)
The oxidation state of Al is +3 in both reactants and products and that of hydrogen it is +1 and oxygen -1
5) E0 Zn+2 = 0.76
E0Fe+2 = 0.44
So E0cell = E0cathode - E0anode
E0cell = -0.44 - (-0.76) = 0.32 V
Ecell = E0cell - (0.0592 / 2 ) log[Zn+2 ] / [Fe+2]
Ecell = 0.32 - 0.0296 log (0.005 / 0.05) = 0.32 - 0.0296 (-1) = 0.3496 V
6) We know that
G° = -RT ln K = -2.303 RT log K
K = 1.7x107
so log K = 7.23
G° = - 2.303 X 8.314 X 298 X 7.23 = - 41253.28 Joules / mole = -41.253 KJ / mole (approx = 41 KJ / mole)
7)
We know that delta G NOCl = 66.30 kJ/mol
so Delta G of reaction = 2X Delta G of NOCl - (Delta G Cl2 +2X delta G of NO)
Delta G of reaction = 2X66.3 - ( 0+ 2X 86.6) = 132.6 - 173.2 = -40.6 KJ ( approx -41)
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