# Question #454d0

May 26, 2017

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:

$\Delta G = \Delta H - T \times \Delta S$
With:
$G$=Gibbs free energy ($\frac{J}{m o l}$)
$H$=Enthalpy ($\frac{J}{m o l}$)
$T$=Temperature ($K$)
$S$=Entropy ($\frac{J}{m o l \times K}$)

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 $G = 0$.

$0 > 150 , 000 - T \times - 2550$
$0 > 150 , 000 + 2550 \times T$
$2550 \times T < - 150 , 000$
$T < - 58.8$

As you can see, the temperature should be $- 58.8 K$, 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).
$\Delta H - T \times \Delta S$ gives
$x - T \times - y$
$x + y T$
Since x, y and T are both positive, this reaction will never have a Gibbs free energy below 0, and therefore cannot be spontaneous.