A- Consider the reaction 4 HCl(g) + O 2 (g) = 2 H 2 O(g) + 2 Cl 2 (g) Using the

Question

A-
Consider the reaction

4HCl(g) + O2(g)=2H2O(g) + 2Cl2(g)

Using the standard thermodynamic data in the tables linked above, calculate delta G for this reaction at 298.15K if the pressure of each gas is 21.79 mm Hg.

ANSWER: ( ) kJ/mol

B- Consider the reaction

N2(g) + 2O2(g)=2NO2(g)

Use the standard thermodynamic data in the tables linked above. Calculate delta G for this reaction at 298.15K if the pressure of NO2(g) is reduced to 17.51mm Hg, while the pressures of N2(g) and O2(g) remain at 1 atm.

ANSWER: ( ) kJ/mol

C-
Consider the reaction

2N2(g) + O2(g)=2N2O(g)

Using the standard thermodynamic data in the tables linked above, calculate delta Grxn for this reaction at 298.15K if the pressure of each gas is 33.33 mm Hg.

ANSWER: ( ) kJ/mol

Explanation / Answer

A. 4HCl(g) + O2(g)=2H2O(g) + 2Cl2(g)

298.15K if the pressure of each gas is 21.79 mm Hg.

Using the equation G= RTln P/Po

Where R –gas constant = 8.314JK-1mol-1

T-temperature K

P- final pressure of the reaction P= 2xPH2O+ 2x P Cl2= (2x 21.79)+(2x 21.79)= 87.16mmHg

Po initial pressure of the reaction Po = 4x PHCl+P O2= (4x 21.79+ 21.79) mmHg = 108.95mmHg

Since

G= 8.314JK-1mol-1 x 298.15 K ln(87.16mmHg/108.95 mmHg)

      = 2478.81Jmol-1 x ln 0.8

      = 2478.81Jmol-1 x -0.2231

      = -553.13J/mol

     G = 0.5531KJ/mol

B.N2(g) + 2O2(g)=2NO2(g)

298.15K if the pressure of NO2(g) is reduced to 17.51mm Hg, while the pressures of N2(g) and O2(g) remain at 1 atm. Using the equation G= RTln P/Po

Where R –gas constant = 8.314JK-1mol-1

T-temperature K

P- final pressure of the reaction P= PNO2= 17.51mmHg

Po initial pressure of the reaction Po = PN2+P O2= (760+760mmHg) = 1520mmHg   1atm= 760mmHg

Since

G= 8.314JK-1mol-1 x 298.15 K ln(17.51 mmHg/1520mmHg)

      = 2478.81Jmol-1 x ln0.01151

      = 2478.81Jmol-1 x -2.1619

      = 5359.07   Jmol-1

     G = 5.359KJ/mol    

C. 2N2(g) + O2(g)=2N2O(g)

298.15K if the pressure of each gas is 33.33 mm Hg.               

Where R –gas constant = 8.314JK-1mol-1

T-temperature K

P- final pressure of the reaction P=2x PN2O= 2x 33.33mmHg = 66.66mmHg

Po initial pressure of the reaction Po = 2xPN2+P O2= (66.66+33.33mmHg) = 99.99mmHg  

Since

G= 8.314JK-1mol-1 x 298.15 K ln(66.66 mmHg/99.99mmHg)

      = 2478.81Jmol-1 x ln0.666

      = 2478.81Jmol-1 x -0.4064

      = 1007.55   Jmol-1

     G = 1.0075KJ/mol    

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