Questions 41-43 are based on the following description. When ignited, solid ammo
ID: 695546 • Letter: Q
Question
Questions 41-43 are based on the following description. When ignited, solid ammonium dichromate decomposes in a fiery display. This is the reaction for a "volcano" demonstration. The decomposition produces nitrogen gas, water vapor, and chromium(III) oxide. The temperature is constant at 25°C Substance AH (kJ/mol) So (kJ/mol K) -1147 -242 Cr20s(g) H20(0) 0.08118 0.1187 0.1915 0.1137 N2(g) NHCr0+) 5 -22.5 -41 . Determine Sunio (in kJ/mol K). A) 7.66 B) 6.39 C) 84.3 D) 5.22 E) 6.03 42. Determine aso reaction (in kJ/mol K). A) 0.2777 B) 0.8612 C) 0.7475 D) 0.6338 E) 0.1590 43. Determine Go (in kJ/mol). A) -191.4 B) -2281.4 C)-38.9 D) 1903.6 E) -1555.4Explanation / Answer
(NH4)2Cr2O7(s) ------> N2(g) + H2O(l) + Cr2O3(g)
The balanced equation is
(NH4)2Cr2O7(s) ------> N2(g) + 4H2O(l) + Cr2O3(g)
41.
dH0rxn = Sum of dH0(products) - Sum of dH0(reactants)
= [ 0 + 4 (- 242) + ( - 1147) ] – (- 22.5)
= [ 0 - 968 - 1147 ] + 22.5)
= - 2115 + 22.5
= - 2092.5 kJ/mol
dS0rxn = Sum of dS0(products) - Sum of dS0(reactants)
= [ 0.1915 + 4(0.1187) + 0.08118] – (0.1137)
= 0.63378
= 0.6338 kJ/mol K
Now,
dS0Universe = dS0system + dS0surroundings
System is the reaction
dS0system = dS0rxn = 0.6338
dS0surrounding = - dH0rxn / T
= -(-2092.5) kJ/mol / 298 K
= 7.02 kJ/mol K
dS0Universe = dS0system + dS0surroundings
= 0.6338 kJ/mol K + 7.02 kJ/mol K
= 7.655 kJ/mol K
= 7.66 kJ/mol K
Answer is option (A) 7.66
42.
dS0 = Sum of dS0(products) - Sum of dS0(reactants)
= [ 0.1915 + 4(0.1187) + 0.08118] – (0.1137)
= 0.63378
= 0.6338 kJ/mol K
Answer is option (D) 0.6338
43.
dG0 = dH0 =- T dS0
= (- 2092.5 kJ/mol) – 298 K (0.6338 kJ/mol K)
= (- 2092.5 kJ/mol) – 188.9 kJ/mol
= - 2281.4 kJ/mol
Answer is option (B) – 2281.4
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