Practice Set of Thermodynamics Problems

For doing questions of type one to seven, usually the value of change in
enthalpy dH and the change in entropy dS value is given in the question
itself. I have got the standard values for the first reaction but not
the others. So I have outlined the method for all the problems. You just
need to put in the values of dH and dS which I suppose you might have as
a table or something supplied along with this question set.

Please feel free to contact me for any clarfications.

 (10 points)

Top of Form

Calculate the minimum temperature above which the following reaction
will be product-favored (spontaneous).

2 Ag2O(s) -------( 4Ag(s) + 02(g)







a.

298 K  



b.

467.8 K  



c.

0.002137 K  



d.

2.137 K  



Answer is A

We know that, dG = dH – TdS

For the above reaction, dH = 30.56kJ/mol and dS = 66.0 J/Kmol

For the reaction to be spontaneous,

dG must be negative.

That is TdS > dH

( T *66 > 305060

( T > 30560/66 = 463 K

( T > 463 K

So the answer is B.

(We must know the value of enthalpy change and entropy change to do
these types of problems. I have taken the standard values for he above
reaction. Please check whether you took different values from mine for
dH and dS.)





 (10 points)

Top of Form



a.

4.16 K  



b.

3645 K  



c.

298 K  



d.

4160 K  

Calculate the minimum temperature above which the following reaction
will be product-favored (spontaneous).

2Cr2 03(s) --( 4Cr(s) + 3O2(g)

,









Answer is A

We know that, dG = dH – TdS

For the above reaction, dH = x J/mol and dS = y J/Kmol

For the reaction to be spontaneous,

dG must be negative.

That is TdS > dH

( T *y > x

( T > x/y

 (10 points)

Top of Form

Calculate the minimum temperature above which the following reaction
will be product-favored (spontaneous).

2NiO(s) --( 2Ni(s) + O2(g)







a.

2.538 K  



b.

0.0003940 K  



c.

2538 K  



d.

4357 K  



Answer is B

We know that, dG = dH – TdS

For the above reaction, dH = x J/mol and dS = y J/Kmol

For the reaction to be spontaneous,

dG must be negative.

That is TdS > dH

( T *y > x

( T > x/y

 (10 points)

Top of Form

Calculate the minimum temperature above which the following reaction
will be product-favored (spontaneous).



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S/Tables/ThermoTable.htm','new_window','width=480,height=360,menubar=0,t
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P4O10(s) -( P4(s,white) + 5O2(g)





a.

310.45 K  



b.

0.00032212 K  



c.

3104.5 K  



d.

298 K  



Answer is B

We know that, dG = dH – TdS

For the above reaction, dH = x J/mol and dS = y J/Kmol

For the reaction to be spontaneous,

dG must be negative.

That is TdS > dH

( T *y > x

( T > x/y



 (10 points)

Top of Form

Calculate the minimum temperature above which the following reaction
will be product-favored (spontaneous).



HYPERLINK
"javascript:void(doWindowOpen=window.open('/CHE112C11_05M_WiltFlemon/CTM
S/Tables/ThermoTable.htm','new_window','width=480,height=360,menubar=0,t
oolbar=0,scrollbars=1,status=1,location=0,resizable=1,resizable=1',0))"
2Al2O3(s) -( 4Al(s) + 3O2(g)





a.

5350 K  



b.

5.35 K  



c.

298 K  



d.

0.000187 K  

Answer is C

We know that, dG = dH – TdS

For the above reaction, dH = x J/mol and dS = y J/Kmol

For the reaction to be spontaneous,

dG must be negative.

That is TdS > dH

( T *y > x

( T > x/y

 (10 points)

Top of Form

Calculate the minimum temperature above which the following reaction
will be product-favored (spontaneous).

2Ca0(s) -( 2Ca(s) + O2(g)







a.

298 K  



b.

6.093 K  



c.

6093 K  



d.

7643 K  



Answer is D

We know that, dG = dH – TdS

For the above reaction, dH = x J/mol and dS = y J/Kmol

For the reaction to be spontaneous,

dG must be negative.

That is TdS > dH

( T *y > x

( T > x/y

 

Top of Form

Calculate the minimum temperature above which the following reaction
will be product-favored (spontaneous).

Fe3O4(s) ---( 3Fe(s) + 2O2(g)







a.

3.221 K  



b.

4105 K  



c.

3221 K  



d.

0.0003105 K  





Answer is A

We know that, dG = dH – TdS

For the above reaction, dH = x J/mol and dS = y J/Kmol

For the reaction to be spontaneous,

dG must be negative.

That is TdS > dH

( T *y > x

( T > x/y





Question 8 

Top of Form

The thermodynamic equilibrium constant for the following reaction is
0.15 at 1227°C.

2 SO2(g) + O2(g) <=>2 SO3(g)

Check the box for each true statement.





a.

The reaction is product-favored.  



b.

The reaction is reactant-favored.  



c.

G°is less than zero.  



d.

G° is greater than zero.  



e.

G° is equal to zero.  

B,C

We know that, G0 = -RT log K = - R*2.303*(1227+273)*log 0.15 = 2846R

So, we can see that change in gibbs free energy is positive (d). Hence
the reaction is reactant favoured (b). Hence the answers are b and d





Question 9 

Top of Form

The thermodynamic equilibrium constant for the following reaction is
2.31 x 10-3 at 1073K.

2HI(g) <=>H2(g) + I2(g)

Check the box for each true statement.





a.

The reaction is product-favored.  



b.

The reaction is reactant-favored.  



c.

G° is less than zero.  



d.

G° is greater than zero.  



e.

G° is equal to zero.  



Answer is A,E

We know that, G0 = -RT*2.303* log K = - R*2.303*(1073+273)*log 2.31 x
10-3 = 8172R

So, we can see that change in gibbs free energy is positive (d). Hence
the reaction is reactant favoured (b). Hence the answers are b and d





 (10 points)

Top of Form

The thermodynamic equilibrium constant for the following reaction is
0.1764 at 1500°C.

CO(g) + 3H2(g) <=>CH4(g) + H2O(g)

Check the box for each true statement.





a.

The reaction is product-favored.  



b.

The reaction is reactant-favored.  



c.

G° is less than zero.  



d.

G° is greater than zero.  



e.

G° is equal to zero.  

Bottom of Form



Answer is D, A

When G is greater than zero, the reaction is nonspontaneous in the
forward direction.

We know that, G0 = -RT*2.303* log K = - R*2.303*(1500+273)*log 0.1764=
3076R

So, we can see that change in gibbs free energy is positive (d). Hence
the reaction is reactant favoured (b). Hence the answers are b and d

Please feel free to contact me through BM if you need any
clarification.

Best wishes

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