Delta H is 20.1 kJ/mol and Delta S is 45.9 J/(mol-k). Assuming these values change very little with temperature, over what temperature range is the reaction spontaneous in the forward direction?

Is the reaction spontaneous for temperatures less than or greater than the solved temperature?

here's what i did but i doubt it's correct
-0.0001 = ∆H -T∆S
-0.0001 = (20.1 kJ/mol) - T(45.9 J/mol-k)
-20.1001 = -45.9T
the reaction is spontaneous for temperatures greater than T = 0.4379 K?

i used -0.0001 to make Delta G spontaneous

Right ball park but out at first base.

One USUALLY sets dG = 0 although your -0.0001 is not right ONLY for those values between zero and -0.0001 (of course the mathematicians probably will say that's an infinite number of dG values). Next, and a much larger error is you mixed units. DH is given in kJ/mol and dS in J/mol. You need to change one of them. Therefore, your T value is too small by a factor of 1000. It should be 437.9 K.

Dang.. thanks DrBob!

To determine the temperature range over which the reaction is spontaneous in the forward direction, we can use the equation for Gibbs free energy (ΔG) in terms of enthalpy (ΔH) and entropy (ΔS):

ΔG = ΔH - TΔS

For the reaction to be spontaneous in the forward direction, ΔG must be negative.

So, if we set ΔG equal to zero and solve for temperature (T):

0 = ΔH - TΔS

Rearranging the equation to solve for T:

T = ΔH / ΔS

Now substituting the given values of ΔH and ΔS:

T = (20.1 kJ/mol) / (45.9 J/(mol-K))

Converting 20.1 kJ to J:

T = (20.1 × 10^3 J/mol) / (45.9 J/(mol-K))

Simplifying:

T = 438.15 K

Therefore, the reaction is spontaneous in the forward direction for temperatures greater than 438.15 K.

Regarding the second part of your question, the reaction is spontaneous for temperatures greater than the calculated temperature (T > 438.15 K). This means that the reaction would not be spontaneous for temperatures less than 438.15 K.

To determine the temperature range over which the reaction is spontaneous in the forward direction, we can use the equation for the Gibbs free energy change (∆G) in terms of enthalpy change (∆H) and entropy change (∆S):

∆G = ∆H - T∆S

In this case, ∆H = 20.1 kJ/mol (given) and ∆S = 45.9 J/(mol-K) (given). Since the units of ∆H and ∆S are different, we need to convert them to have the same units.

∆H = 20.1 kJ/mol = 20,100 J/mol
∆S = 45.9 J/(mol-K)

Now we can substitute these values into the equation:

∆G = 20,100 J/mol - T * 45.9 J/(mol-K)

To determine the temperature range where the reaction is spontaneous in the forward direction, we need to find the temperature at which ∆G is negative (∆G < 0). This means the reaction is spontaneous and will proceed in the forward direction.

So, we need to solve the equation:

20,100 J/mol - T * 45.9 J/(mol-K) < 0

Simplifying the inequality:

20,100 J/mol < T * 45.9 J/(mol-K)

Dividing both sides of the inequality by 45.9 J/(mol-K):

20,100 J/mol / 45.9 J/(mol-K) < T

438.6 K < T

Therefore, the reaction is spontaneous in the forward direction for temperatures greater than 438.6 K.