tif ure 5. Single Correct 00 points It was discussed in class that the Haber Pro
ID: 592936 • Letter: T
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tif ure 5. Single Correct 00 points It was discussed in class that the Haber Process is the major industrial use of hydrogen in the production of ammonia according to the following reaction 3H2 (g) . N2 (g) 2NH3 (g) Chemical Species AH (kJ/mol S (J/mol)K AG(kJ/mol) 192 193 17 Use the tabular data above to answer the following and place the letter corresponding to the answer in the blank provided. Show all work. -Calculate H. for the Haber process reaction at 25K. A. .92.0 kJ B. 10.580 kJ C. -121.315 kJ D. 21.865 kJ E. 155.170 kJ Calculate AS for the Haber process reaction at 25oC. A. 809.09 J/K B. -89.39 J/K C. 453.78 J/K D. -199.00 J/K E. 1164.40 J/K Calculate G ° for the Haber process reaction at 25°C. A. -33x10s kJ B. 98.7 k C. -33 k D. 35 kJ E. 0 -Is the reaction spontaneous at standard conditions? A. Yes B. No At what temperature is the reaction spontaneous at standard conditions, assuming changes in enthalpy and entropy change do not depend upon temperature. A. At T 460 K C. At T = 298 K D. At T=0KExplanation / Answer
1)
we have:
Hof(H2(g)) = 0.0 KJ/mol
Hof(N2(g)) = 0.0 KJ/mol
Hof(NH3(g)) = -46.0 KJ/mol
we have the Balanced chemical equation as:
3 H2(g) + N2(g) ---> 2 NH3(g)
deltaHo rxn = 2*Hof(NH3(g)) - 3*Hof( H2(g)) - 1*Hof(N2(g))
deltaHo rxn = 2*(-46.0) - 3*(0.0) - 1*(0.0)
deltaHo rxn = -92 KJ
Answer: A
2)
we have:
Sof(H2(g)) = 131.0 J/mol.K
Sof(N2(g)) = 192.0 J/mol.K
Sof(NH3(g)) = 193.0 J/mol.K
we have the Balanced chemical equation as:
3 H2(g) + N2(g) ---> 2 NH3(g)
deltaSo rxn = 2*Sof(NH3(g)) - 3*Sof( H2(g)) - 1*Sof(N2(g))
deltaSo rxn = 2*(193.0) - 3*(131.0) - 1*(192.0)
deltaSo rxn = -199 J/K
Answer: D
3)
we have:
Gof(H2(g)) = 0.0 KJ/mol
Gof(N2(g)) = 0.0 KJ/mol
Gof(NH3(g)) = -17.0 KJ/mol
we have the Balanced chemical equation as:
3 H2(g) + N2(g) ---> 2 NH3(g)
deltaGo rxn = 2*Gof(NH3(g)) - 3*Gof( H2(g)) - 1*Gof(N2(g))
deltaGo rxn = 2*(-17.0) - 3*(0.0) - 1*(0.0)
deltaGo rxn = -34 KJ
Answer: C
4)
Since delta G is negative, the reaction is spontaneous
Answer: yes
5)
deltaHo = -92.0 KJ/mol
deltaSo = -199 J/mol.K
= -0.199 KJ/mol.K
we have below equation to be used:
deltaGo = deltaHo - T*deltaSo
for reaction to be spontaneous, deltaGo should be negative
that is deltaGo<0
since deltaGo = deltaHo - T*deltaSo
so, deltaHo - T*deltaSo < 0
-92.0- T *-0.199 < 0
T *0.199 < 92.0
T < 462 K
Answer: A
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