Question

From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 28°C: reaction A: ΔH = 10.5 kJ/mol, ΔS = 30.0 J/K·mol; spontaneous nonspontaneous impossible to tell reaction B: ΔH = 1.8 kJ/mol, ΔS = −113 J/K · mol, spontaneous nonspontaneous impossible to tell If either of the reactions is nonspontaneous, can it(they) become spontaneous? yes, reaction A can become spontaneous yes, reaction B can become spontaneous yes, both reactions can become spontaneous no, neither reaction can become spontaneous If either of the reactions is nonspontaneous but can become spontaneous, at what temperature might it become spontaneous? °C

Answer 1

Answer: A:

ΔH = 10.5 kJ/mol, ΔS = 30.0 J/K·mol is non-spontaneous.

ΔH = 1.8 kJ/mol, ΔS = −113 J/K · mol is non-spontaneous.

Reaction A can become spontaneous

Reaction A is spontaneous at 76.85 °C

Explanation:

It's helpful to memorize that if:

-ΔS is greater than 0 and ΔH is less than 0; its spontaneous at all temperatures.

-ΔS is less than 0 and ΔH is greater than 0; its non-spontaneous at all temperature.

-ΔS is greater than 0 and ΔH is greater than 0; its spontaneous at high temperatures and non-spontaneous at low temperatures.

-ΔS is less than 0 and ΔH is less than 0; its spontaneous at low temperatures and non-spontaneous at high temperatures.

This comes from the equation ΔG=ΔH-TΔS

where ΔG is Gibbs free energy, ΔH is enthalpy, T is temperature (in Kelvin), and ΔS is entropy.

Without getting too in depth as to what each of those mean (you could take an entire class on entropy alone), the temperature at which the spontaneity changes is equal to ΔG (Gibbs free energy) at 0.

So take the above equation and set ΔG = 0, and rearrange the equation to solve for T.

ΔG=ΔH-TΔS

0=ΔH-TΔS

add TΔS to the other side

TΔS=ΔH

divide the right side by ΔS to find T (temperature)

T=ΔH/ΔS

Now we can find the temperature that the first reaction would occur at spontaneously.

We need to make sure that we have the same units for ΔH and ΔS, so divide 30 by 1000 to convert J/Kmol into kJ/kmol so that we have kJ for ΔH and ΔS.

30/1000 = 0.03 kJ

Plug in the values for the modified equation T=ΔH/ΔS

10.5 kJ/0.03 kJ = 350 K

The temperature is in Kelvin, so subtract 273.15 to convert it into Celsius

350-273.15 = 76.85 °C

Answer 2

Answer:

Reaction A is nonspontaneous and it can become spontaneous above 77°C.

Reaction B is nonspontaneous and it cannot become spontaneous.

Explanation:

To determine if a reaction is spontaneous, we need to consider the Gibbs free energy (ΔG).

• If ΔG < 0, the reaction is spontaneous.

• If ΔG > 0, the reaction is nonspontaneous.

ΔG can be calculated using the following expression.

ΔG = ΔH - T.ΔS

where,

ΔH: enthalpy of reaction

T: absolute temperature

ΔS: entropy of reaction

From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 28°C (301 K):

Reaction A: ΔH = 10.5 kJ/mol, ΔS = 30.0 J/K·mol;

ΔG = ΔH - T.ΔS = 10.5 × 10³ J/mol - 301K . 30.0 J/K.mol = 1470 J/mol

Since ΔG > 0, the reaction is nonspontaneous.

It can become spontaneous when ΔG < 0,

ΔH - T.ΔS < 0

ΔH < T.ΔS

T > ΔH / ΔS = (10.5 × 10³ J/mol)/(30.0 J/K.mol) = 350 K

The reaction will become spontaneous above 350 K (77 °C)

Reaction B: ΔH = 1.8 kJ/mol, ΔS = −113 J/K · mol

ΔG = ΔH - T.ΔS = 1.8 × 10³ J/mol - 301K . (-113 J/K.mol) = 3.58 × 10⁴ J/mol

Since ΔG > 0, the reaction is nonspontaneous.

Under no conditions can ΔG be negative, so the reaction cannot be spontaneous.