If a reaction has a change of H of -120.0 kcals/mole and is spontaneous for all temperatures less than 200.0K, what is its ΔS?
dG = dH - TdS
Set dG = 0, T is 200 and dH is -120.
Solve for dS
Then check it to see that temperatures below 200 have dG = negative and temperatures above 200 K have dG = positive.
To determine the change in entropy (ΔS) of a reaction, you can use the equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
T is the temperature in Kelvin
ΔS is the change in entropy
In this case, we know that the reaction is spontaneous for all temperatures below 200.0K. This indicates that the change in Gibbs free energy (ΔG) is negative for temperatures below 200.0K.
Since ΔG = ΔH - TΔS, we can rearrange the equation to solve for ΔS:
ΔS = (ΔH - ΔG) / T
Since the reaction is spontaneous at all temperatures below 200.0K, we know that ΔG is negative. Therefore, ΔS can be calculated as follows:
ΔS = (ΔH - (-ΔG)) / T
ΔS = (ΔH + ΔG) / T
Given that ΔH = -120.0 kcals/mole and ΔG is negative, we can substitute these values into the equation:
ΔS = (-120.0 kcals/mole + ΔG) / T
However, since we don't have the specific value of ΔG, we cannot calculate the exact value of ΔS without that information.
To find the value of ΔS (the change in entropy) for this reaction, we can use the Gibbs free energy equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in free energy
ΔH is the change in enthalpy
T is the temperature in Kelvin
ΔS is the change in entropy
We are given that ΔH is -120.0 kcal/mole (the change in enthalpy), and the reaction is spontaneous for all temperatures less than 200.0 K. This means that ΔG is negative for temperatures less than 200.0 K.
Since the reaction is spontaneous, we can assume that the value of ΔG is zero at this temperature. So the equation becomes:
0 = -120.0 - (200.0 * ΔS)
We can rearrange the equation to solve for ΔS:
ΔS = -120.0 / 200.0
Calculating this expression, we find:
ΔS = -0.6 kcal/(mol*K)
Therefore, the change in entropy (ΔS) for this reaction is -0.6 kcal/(mol*K).