Question

Using the data in Appendix C in the textbook and given the pressures listed, calculate ΔG for each of the following reactions at 298 K.

Part A

N2(g)+3H2(g)→2NH3(g)
  

Express your answer using two significant figures. If your answer is greater than 10100, express it in terms of the base of the natural logarithm using two decimal places: for example, exp(200.00).

Kp =

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Part B

N2(g)+3H2(g)→2NH3(g)
PN2 = 3.6 atm , PH2 = 5.8 atm , PNH3 = 1.6 atm

Express your answer using three significant figures.

ΔG =

  kJ  

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Part C

2N2H4(g)+2NO2(g)→3N2(g)+4H2O(g)
  

Express your answer using two significant figures. If your answer is greater than 10100, express it in terms of the base 10 logarithm using two decimal places: for example, 10 ^(200.00)

Kp =

Part D

2N2H4(g)+2NO2(g)→3N2(g)+4H2O(g)
PN2H4=PNO2=4.5×10−2atm, PN2 = 1.9 atm , PH2O = 2 atm

Express your answer using three significant figures.

ΔG =

  kJ  

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Part E

N2H4(g)→N2(g)+2H2(g)
  

Express your answer using two significant figures. If your answer is greater than 10100, express it in terms of the base of the natural logarithm using two decimal places: for example, exp(200.00).

Kp =

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Part F

N2H4(g)→N2(g)+2H2(g)
PN2H4 = 0.1 atm , PN2 = 2.6 atm , PH2 = 8.1 atm

Express your answer using four significant figures.

ΔG =

  kJ  

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Answer 1

N₂(g)+3H₂(g)→2NH₃(g)

PART A:
K_p = P_NH3² / P_N2 * P_H2³ = 1.6²/ (3.6 * 5.8³) = 3.64 * 10^-3

PART B:

ΔG = - R * T * ln (K_p) = - 8.314 * 298 * ln ( 3.64 * 10^-3) = 13910 J/ mole = 13.9 kJ/ mole

2 N₂H₄(g)+2NO₂(g)→ 3 N₂(g)+ 4 H₂O(g)

PART C:

Kp = P_N2³ * P_H2O⁴/ P_N2H4² * P_NO2² = 1.9³ * 2⁴ / ( 4.5×10⁻² )² * ( 4.5×10⁻² )² = 54195 = 10^4.73

PART D:

ΔG = - R * T * ln (K_p) = - 8.314 * 298 * ln ( 54195) = - 27006 J/ mole = -27.0 kJ/ mole

N₂H₄(g)→ N₂(g) + 2 H₂(g)

PART E:

Kp = P_N2 * P_H2²/ P_N2H4 = 2.6 * ( 8.1)²/ 0.1 = 1705.86

PART F:

ΔG = - R * T * ln (K_p) = - 8.314 * 298 * ln ( 1705.86) =- 18438 J/ mole = - 18.44 kJ/ mole