Question

Question #454d0

Answer 1

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.