3. (20 Points) Thermodynamics \" = + 260.8 kJ AS= + 278.5 J/K Consider the react

Question

3. (20 Points) Thermodynamics " = + 260.8 kJ AS= + 278.5 J/K Consider the reaction, 2 NF3 (g)2N2(g) + 3 F2 (g) alement below is true for the system under standard conditions (all gases 1 atm) at 25*C? A) The system is at equilibrium B) The forward reaction wil proceed spontaneously C) Th e reverse reaction will proceed spontaneously D) The entropy change for the universe is positive ii. As the system precedes in the forward direction (as written) which statement best applies A) There is a net increase in the disorder of the system B) Heat energy is released C) The free energy change is zero D) All of the above ii At what temperature will the system be at equilibrium, when all gases are present at 1 atm (standard pressure conditions)? (6) 6600 C)25 D) 940 °C A)-270 A)-270 °C alahe evfthe eguilibrium constant for the reaction above at 500.0c

Explanation / Answer

i.

the correct answer is c)

the reverse reaction will proceed spontaneously

where G is: negative, the process is spontaneous

dG = dH -TdS

Here dH = 260.8 Kj = 260.8*10^3 J

dS = 278.5J /K

T = 298 K because dH and dS are in standard conditions

Then;

dG = dH -TdS

dG = [260800 – 298 * 278.5 ]J

dG = [260800 – 82993 ]J

dG = [+177807 ]J or177.8 KJ

and for reverse reaction dG = - 177807 J or - 177.8 KJ

ii.

Consider the reaction

2 NF3 = 2N2 + 3F2

the correct answer is A)

because there are number of gaseous molecule swill increases

iii. at equilibrium dG = 0 then

Then;

dG = dH -TdS

0 = [260800 – Tin K * 278.5 ]

T in K = 260800 / 278.5 =936.5 K

And now change the unit from Kelvin to degree Celsius as follows:

degree Celsius =936.5 K – 273.15

= 663.3 C

= 660 C

Thus the correct answer is B

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