Practice Problem 8
Predict whether the following reaction is still spontaneous at 500°C:
N2(g) + 3 H2(g) 2 NH3(g)
Assume that the values of Ho and S° used in Practice Problem 7 are still valid at this temperature.
Before we can decide whether the reaction is still spontaneous we need to calculate the temperature of the kelvin scale:
TK = 500o C + 273 = 773 K
We then multiply the entropy term by this temperature and subtract this quantity from the enthalpy term:
Go773 = Ho298 - T So298
= 92,220 J - (773 K x -198.75 J/K)
= 92,220 J - (-153,600 J)
= 61,380 J
Go = 61.4 kJ
Because the entropy term becomes larger as the temperature increases, the reaction changes from one which is favorable at low temperatures to one that is unfavorable at high temperatures.