1) Calculate the standard entropy change for the following reaction. 2N 2 (g) +
ID: 1006432 • Letter: 1
Question
1)
Calculate the standard entropy change for the following reaction.
2N2(g)
+ 6H2(g)
?
4NH3(g)
191.5
130.6
192.3
794.8 J/K?mol-rxn
–129.8 J/K?mol-rxn
–794.8 J/K?mol-rxn
–737.2 J/K?mol-rxn
737.2 J/K?mol-rxn
-----------
2)
For the reaction given below, ?H0 = ?1516 kJ at 25°C and ?S0 = ?432.8 J/K at 25°C. This reaction is spontaneous ____.
at no temperatures
only below a certain temperature
at all temperatures
only above a certain temperature
cannot tell from the information available
-----------
3)
Consider the following half-reactions:
Which of the above metals or metal ions will oxidize Pb(s)?
Cu2+(aq) and Fe2+(aq)
Ag(s) and Cu(s)
Ag+(aq) and Cu2+(aq)
Fe(s) and Al(s)
Fe2+(aq) and Al3+(aq)
-----------
4)
What is the balanced spontaneous reaction and standard cell potential of an electrochemical cell constructed from half cells with the following half reactions?
Pt2+(aq) + 2e? ? Pt(s) E° = 1.180 V
Pb2+(aq) + 2e? ? Pb(s) E° = –0.130 V
2N2(g)
+ 6H2(g)
?
4NH3(g)
(J/mol?K)191.5
130.6
192.3
Explanation / Answer
1)
consider the given reaction
2 N2 + 6 H2 --> 4 NH3
we know that
dSo rxn = So products - So reactants
so
dSo rxn = ( 4 x So NH3) - ( 2 x So N2) - ( 6 x So H2)
dSo rxn = ( 4 x 192.3 ) - ( 2 x 191.5) - ( 6 x 130.6)
dSo rxn = -397.4
2)
we know that
dG = dHo - TdSo
for a reaction to be spontaneous , dG < 0
consider
dG <
dHo - TdSo < 0
TdSo > dHo
T < dHo / dSo
T < -1516 x 1000 / -432.8
T < 3502
so
the reaction is spontaneous for temperatures below 3502 K
so
the answer is
only below a certain temperature
3)
the required metal should oxidize Pb (s)
So
the required metal should undergo reduction
we know that
for a metal to undergo reduction , it should act as a cathode
now
to be a cathode , the metal should have higher reduction potential than anode (Pb)
from the given table
we can see that
Cu+2 and Ag+ have higher reduction potentials
so
the answer is Ag+ (aq) and Cu+2 (aq)
4)
we know that
half cell with higher reduction potential is cathode
also
reduction takes place at cathode
so
Cathode : reduction
Pt+2 + 2e- --> Pt
half cell with lower reduction potential is anode
also
oxidation takes place at anode
so
anode : oxidation
Pb (s) ---> Pb+2 + 2e-
So
the overall reaction is
Pt+2 (aq) + Pb (s) ---> Pb+2 (aq) + Pt (s)
now
EO cell = Eo cathode - Eo anode
Eo cell = 1.18 - ( -0.13)
Eo cell = 1.31 V
so
the answer is
Pt+2 (aq) + Pb (s) ---> Pb+2 (aq) + Pt (s) --> Eo cell = 1.31 V
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