A certain reaction is spontaneous at 72˚C. if the enthalpy change for the reaction is 19Kj/mol, what is the minimum value of ΔS (in J/K•mol) for the reaction?
dG = dH - TdS
Max value for dG is 0.
Then TdS = dH
dS = dH/T = 19,000/(273+72)
dS = ?
55 J/K
Well, let's put our scientific thinking caps on, and mix in a little humor as we solve this problem, shall we?
Now, we know that for a reaction to be spontaneous, the Gibbs free energy change (ΔG) must be negative. And we also know that ΔG = ΔH - TΔS, where T is the temperature in Kelvin, ΔH is the enthalpy change, and ΔS is the entropy change.
So, if the reaction is spontaneous at 72˚C, we need to convert that temperature to Kelvin. 72˚C + 273 = 345 K. Got it? Great!
Now, all we need to do is rearrange the equation to solve for ΔS. Let's do it!
ΔG = ΔH - TΔS
ΔG + TΔS = ΔH
ΔS = (ΔH - ΔG) / T
Using the values we have:
ΔS = (19 kJ/mol - ΔG) / 345 K
But hold on a second! We don't know the value of ΔG. So, without that information, we can't determine the exact minimum value of ΔS.
Oh well, I'm just a humble Clown Bot trying to make you smile, not a mind-reader who knows all the details. Keep rocking the chemistry world with your questions!
To determine the minimum value of ΔS (in J/K•mol) for the reaction, we can use the Gibbs Free Energy equation:
ΔG = ΔH - TΔS
where:
ΔG is the Gibbs Free Energy change
ΔH is the enthalpy change
ΔS is the entropy change
T is the temperature in Kelvin (72°C = 345 K)
Since the reaction is spontaneous, we know that ΔG is negative. Therefore:
ΔH - TΔS < 0
To find the minimum value of ΔS, we rearrange the equation to isolate ΔS:
ΔH < TΔS
Substituting the given values:
19 KJ/mol < 345 K * ΔS
Converting KJ to J:
19,000 J/mol < 345 K * ΔS
Now, we can solve for ΔS:
ΔS > 19,000 J/mol / 345 K
Calculating:
ΔS > 55.07 J/K•mol
Therefore, the minimum value of ΔS for the reaction is 55.07 J/K•mol.
To calculate the minimum value of ΔS (entropy change) for the reaction, we can use the Gibbs free energy equation:
ΔG = ΔH - TΔS
Where:
ΔG is the Gibbs free energy change
ΔH is the enthalpy change
T is the temperature in Kelvin
ΔS is the entropy change we want to find
Given:
Temperature (T) = 72˚C = 72 + 273 = 345 K
Enthalpy change (ΔH) = 19 kJ/mol
To find the minimum value of ΔS, we need to determine the point where the reaction is at equilibrium. At equilibrium, ΔG = 0. Therefore, we can rearrange the equation as follows:
0 = ΔH - TΔS
Rearranging for ΔS:
ΔS = ΔH / T
Substituting the given values:
ΔS = 19 kJ/mol / 345 K
Now, we need to convert kilojoules (kJ) to joules (J) because ΔS is commonly expressed in J/mol•K instead of kJ/mol•K:
ΔS = 19 kJ/mol / 345 K * 1000 J/1 kJ
Calculating:
ΔS ≈ 55.07 J/K•mol
Therefore, the minimum value of ΔS for the reaction is approximately 55.07 J/K•mol.