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Thermodynamics question

2023 · 24 Jan · Shift 1 · Q14
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Thermodynamics question

2023 · 24 Jan · Shift 1 · Q14

JEE MainChemistryThermodynamicsNumerical+4 / −1

For independent process at 300 K

Process ΔH/kJ mol−1\mathrm{\Delta H/kJ~mol^{-1}}ΔH/kJ mol−1 ΔS/J K−1\mathrm{\Delta S/J~K^{-1}}ΔS/J K−1
A −25-25−25 −80-80−80
B −22-22−22 40
C 25 −50-50−50
D 22 20

The number of non-spontaneous process from the following is ‾\underline{\hspace{2cm}}​

Numerical answer
View written solutionFree

Correct answer: 2

We determine spontaneity at constant temperature using Gibbs free energy:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

A process is:

  • spontaneous if ΔG<0\Delta G < 0ΔG<0
  • non-spontaneous if ΔG>0\Delta G > 0ΔG>0

Given T=300 KT = 300\,\text{K}T=300K.

We must convert entropy units from J K−1mol−1\text{J K}^{-1}\text{mol}^{-1}J K−1mol−1 to kJ K−1mol−1\text{kJ K}^{-1}\text{mol}^{-1}kJ K−1mol−1 when needed.

1. Process A

Given:

ΔH=−25 kJ mol−1,ΔS=−80 J K−1mol−1\Delta H = -25\,\text{kJ mol}^{-1}, \qquad \Delta S = -80\,\text{J K}^{-1}\text{mol}^{-1}ΔH=−25kJ mol−1,ΔS=−80J K−1mol−1

Convert entropy term:

ΔS=−0.080 kJ K−1mol−1\Delta S = -0.080\,\text{kJ K}^{-1}\text{mol}^{-1}ΔS=−0.080kJ K−1mol−1

Now,

TΔS=300(−0.080)=−24 kJ mol−1T\Delta S = 300(-0.080) = -24\,\text{kJ mol}^{-1}TΔS=300(−0.080)=−24kJ mol−1

So,

ΔG=−25−(−24)=−1 kJ mol−1\Delta G = -25 - (-24) = -1\,\text{kJ mol}^{-1}ΔG=−25−(−24)=−1kJ mol−1

Thus, process A is spontaneous.

2. Process B

Given:

ΔH=−22 kJ mol−1,ΔS=40 J K−1mol−1\Delta H = -22\,\text{kJ mol}^{-1}, \qquad \Delta S = 40\,\text{J K}^{-1}\text{mol}^{-1}ΔH=−22kJ mol−1,ΔS=40J K−1mol−1

Convert entropy:

ΔS=0.040 kJ K−1mol−1\Delta S = 0.040\,\text{kJ K}^{-1}\text{mol}^{-1}ΔS=0.040kJ K−1mol−1

Then,

TΔS=300(0.040)=12 kJ mol−1T\Delta S = 300(0.040) = 12\,\text{kJ mol}^{-1}TΔS=300(0.040)=12kJ mol−1

So,

ΔG=−22−12=−34 kJ mol−1\Delta G = -22 - 12 = -34\,\text{kJ mol}^{-1}ΔG=−22−12=−34kJ mol−1

Thus, process B is spontaneous.

3. Process C

Given:

ΔH=25 kJ mol−1,ΔS=−50 J K−1mol−1\Delta H = 25\,\text{kJ mol}^{-1}, \qquad \Delta S = -50\,\text{J K}^{-1}\text{mol}^{-1}ΔH=25kJ mol−1,ΔS=−50J K−1mol−1

Convert entropy:

ΔS=−0.050 kJ K−1mol−1\Delta S = -0.050\,\text{kJ K}^{-1}\text{mol}^{-1}ΔS=−0.050kJ K−1mol−1

Then,

TΔS=300(−0.050)=−15 kJ mol−1T\Delta S = 300(-0.050) = -15\,\text{kJ mol}^{-1}TΔS=300(−0.050)=−15kJ mol−1

So,

ΔG=25−(−15)=40 kJ mol−1\Delta G = 25 - (-15) = 40\,\text{kJ mol}^{-1}ΔG=25−(−15)=40kJ mol−1

Thus, process C is non-spontaneous.

4. Process D

Given:

ΔH=22 kJ mol−1,ΔS=20 J K−1mol−1\Delta H = 22\,\text{kJ mol}^{-1}, \qquad \Delta S = 20\,\text{J K}^{-1}\text{mol}^{-1}ΔH=22kJ mol−1,ΔS=20J K−1mol−1

Convert entropy:

ΔS=0.020 kJ K−1mol−1\Delta S = 0.020\,\text{kJ K}^{-1}\text{mol}^{-1}ΔS=0.020kJ K−1mol−1

Then,

TΔS=300(0.020)=6 kJ mol−1T\Delta S = 300(0.020) = 6\,\text{kJ mol}^{-1}TΔS=300(0.020)=6kJ mol−1

So,

ΔG=22−6=16 kJ mol−1\Delta G = 22 - 6 = 16\,\text{kJ mol}^{-1}ΔG=22−6=16kJ mol−1

Thus, process D is non-spontaneous.

5. Count non-spontaneous processes

Non-spontaneous processes are:

  • C
  • D

Hence, the number of non-spontaneous processes is

2\boxed{2}2​
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