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.