A reaction has delta H_rxn= -125 kJ and delta S_rxn= 324 J/K.
At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings?
To find the temperature at which the change in entropy for the reaction is equal to the change in entropy for the surroundings, we can use the equation:
ΔS_rxn = ΔS_surroundings
Given:
ΔH_rxn = -125 kJ
ΔS_rxn = 324 J/K
We know that:
ΔH_rxn = T * ΔS_rxn
where T is the temperature in Kelvin.
Rearranging the equation, we get:
T = ΔH_rxn / ΔS_rxn
Converting the units to kJ and K, we have:
T = (-125 kJ) / (324 J/K)
T = -125,000 J / 324 J/K
T ≈ -385.8 K
It is important to note that you cannot have a negative temperature in Kelvin since Kelvin is an absolute temperature scale. Thus, the change in entropy for the reaction is never equal to the change in entropy for the surroundings.
To determine the temperature at which the change in entropy for the reaction is equal to the change in entropy for the surroundings, you can use the equation:
ΔS_total = ΔS_reaction + ΔS_surroundings
Where
ΔS_total is the total change in entropy,
ΔS_reaction is the change in entropy for the reaction, and
ΔS_surroundings is the change in entropy for the surroundings.
In this case, ΔS_reaction = 324 J/K and we need to find the temperature (T) at which ΔS_total = ΔS_surroundings.
Initially, note that the units for ΔS_reaction and ΔS_surroundings are J/K, while ΔH_rxn is given in kJ. Therefore, we need to convert ΔH_rxn to J before proceeding.
ΔH_rxn = -125 kJ = -125,000 J (-1 kJ = -1000 J)
Now, we can substitute the values into the equation:
ΔS_total = ΔS_reaction + ΔS_surroundings
ΔH_rxn / T = ΔS_reaction + (-ΔH_rxn / T)
Substituting the values we have:
-125,000 J / T = 324 J/K + (-125,000 J / T)
Next, let's simplify the equation:
-125,000 J / T - (-125,000 J / T) = 324 J/K
-250,000 J / T = 324 J/K
Now, to solve for T, let's isolate it on one side of the equation:
-250,000 J / T = 324 J/K
T / -250,000 J = 1 / (324 J/K)
T = -250,000 J / (324 J/K)
T ≈ -771.60 K
Note that the negative sign indicates an illegal temperature since temperature cannot be negative. This suggests that the change in entropy for the reaction is never equal to the change in entropy for the surroundings.
Hence, there is no temperature at which the change in entropy for the reaction is equal to the change in entropy for the surroundings, based on the given values.