Question #454d0

1 Answer
May 26, 2017

Answer:

This reaction will not be spontaneous at any temperature.

Explanation:

A reaction is spontaneous if the Gibbs free energy of that reaction is negative.

The Gibbs free energy can be calculated using:

#DeltaG=DeltaH-TxxDeltaS#
With:
#G#=Gibbs free energy (#J/(mol)#)
#H#=Enthalpy (#J/(mol)#)
#T#=Temperature (#K#)
#S#=Entropy (#J/(molxxK)#)

So we have to get #G<0# to make the reaction spontaneous. Since the other variables are given, we can calculate the temperature range. We fill in the formula for #G=0#.

#0>150,000-Txx-2550#
#0>150,000+2550xxT#
#2550xxT<-150,000#
#T<-58.8#

As you can see, the temperature should be #-58.8K#, which is not possible, the Kelvin scale doesn't contain numbers below 0 because 0 K is the absolute minimum. Therefore we have to conclude that the reaction cannot be spontaneous at any temperature.

Let's take a look at the following example with #H=x# (x= a positive number) and #S=-y# (y is a positive number).
#DeltaH-TxxDeltaS# gives
#x-Txx-y#
#x+yT#
Since x, y and T are both positive, this reaction will never have a Gibbs free energy below 0, and therefore cannot be spontaneous.