Question

Solve:

2 NO2 (g) + O3 (g) → N2O5 (g) + O2 (g)      ∆Hº = -198 kJ mol-1RXN     ∆Sº = -168 J K-1       

Ozone reacts with nitrogen dioxide according to the equation above.

State and explain how the spontaneity of this reaction will vary with increasing temperature.

Substance	∆Hºf (kJ mol -1)
O3 (g)	143
N2O5 (g)	11

Answer 1

suppose lets calculate the ∆Gº at 25 ºC using below formula.

∆Gº = ∆Hº - T∆Sº

convert all in same units so change the ∆Hº in joules

∆Hº = -198000 Jmol-1

T = 273 +25ºC = 298 K

now put all the values in above equation

∆Gº = ∆Hº - T∆Sº

= -198000 - 298 (-168)

= -198000 + 50064

= -147936 J mole^-1 or

= -147.936 kjmole-1

Since ∆Gº is -ve reaction is spontanious

this is the value at 25 ºC

suppose if you increase the temperature

∆Gº =-198000 - T (-168)

in the above equation tight side positive value will increase means ∆Gº value decrease

in the similar way if you increase the temperature more and more ∆Gº value will be positive reaction will be non spontanious

so with increase in temperature this reaction will be non spontanious