For a particular reaction, ΔH° is 20.1 kJ/mol and ΔS° 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?
The reaction is spontaneous for temperatures is greater than T=_____K
dG = dH - TdS
Set dG = 0 and solve for T(in kelvin). Don't forget that dH and dS are not the same unit; i.e., you must convert one of them to the other.
I found approximately 440 K
Well, if we want to determine the temperature range where the reaction is spontaneous in the forward direction, we can use the equation:
ΔG° = ΔH° - TΔS°
So, for the reaction to be spontaneous in the forward direction, ΔG° should be negative. We know that ΔG° = 0 at the equilibrium point, so we can set up the following equation:
0 = 20.1 kJ/mol - T * (45.9 J/(mol·K))
Now, let's convert the units to be consistent:
0 = 20.1 kJ/mol - (1/1000) * T * (45.9 J/(mol·K))
0 = 20.1 - 0.0459 * T
Now, let's solve for T:
0.0459 * T = 20.1
T = 20.1 / 0.0459
T ≈ 437.9 K
So, the reaction is spontaneous in the forward direction for temperatures greater than approximately 437.9 K. But hey, don't forget that these values "change very little with temperature" as mentioned, so this is just an approximation. Keep that in mind, and happy reactions!
To determine the temperature range where the reaction is spontaneous in the forward direction, we need to consider the Gibbs free energy change (∆G°) and the equation for ∆G° in terms of ∆H° and ∆S°:
∆G° = ∆H° - T∆S°
In order for the reaction to be spontaneous in the forward direction, ∆G° must be negative. So we need to find the temperature range where ∆G° is less than zero.
Given:
∆H° = 20.1 kJ/mol = 20,100 J/mol
∆S° = 45.9 J/(mol·K)
∆G° < 0
∆H° - T∆S° < 0
T∆S° > ∆H°
T > ∆H°/∆S°
Plugging in the values:
T > (20,100 J/mol) / (45.9 J/(mol·K))
Calculating this:
T > 438.4 K
Therefore, the reaction is spontaneous in the forward direction for temperatures greater than 438.4 K.
To determine the temperature range over which the reaction is spontaneous in the forward direction, we can use the Gibbs free energy equation:
ΔG° = ΔH° - TΔS°
Where:
ΔG° is the standard Gibbs free energy change
ΔH° is the standard enthalpy change
T is the temperature in Kelvin
ΔS° is the standard entropy change
For a reaction to be spontaneous in the forward direction, ΔG° must be negative (ΔG° < 0).
Rearranging the equation, we have:
ΔG° = -TΔS° + ΔH° < 0
Plugging in the given values:
- T(45.9 J/(mol·K)) + (20.1 kJ/mol) < 0
Now, let's convert 20.1 kJ/mol to J/mol since ΔS° is in J/(mol·K):
- T(45.9 J/(mol·K)) + 20,100 J/mol < 0
To simplify, we'll consider the values in kJ/mol:
- T(0.0459 kJ/(mol·K)) + 20.1 kJ/mol < 0
Now, let's solve for T:
T(0.0459 kJ/(mol·K)) > 20.1 kJ/mol
T > (20.1 kJ/mol) / (0.0459 kJ/(mol·K))
T > 438.79 K
Therefore, the reaction is spontaneous in the forward direction for temperatures greater than 438.79 Kelvin (or ~439 Kelvin).