# Entropy for a Reversible Process

## Trending Questions

**Q.**

5 mole of an ideal gas expand isotherally and irreversibly from a pressure of 10 atm to 1 atm against a contant external pressure of 1 at. Work done at 300 K is:

-15.921 kJ

-11.224 kJ

-110.83 kJ

None of these

**Q.**Acetic acid forms a dimer in the gas phase

The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer. At 25oC, the equilibrium constant for the dimerisation is 1.3×103 (pressure in atm). What is magnitude △S0 for the reaction? Assume that △H does not vary with temperature.

**Q.**1 mole of an ideal gas is allowed to expand isothermally at 27oC until its volume is tripled. Calculate △Ssys and △Suniv under the reversible expansion.

- △Ssys=9.1 J K−1mol−1

- △Suniv=△Ssys=9.1 J K−1mol−1
- △Ssurr=−9.1 J K−1mol−1
- △Suniv=0

**Q.**Standard Free Energy and Equilibrium constant: The change in free energy for a reaction taking place between gaseous reactants and products represented by general equation.

ΔG=ΔG∘+R T lnQP the condition for a system to be at equilibrium is that

ΔG=0 and Qp=KP

Thus at equilibrium

ΔG∘=−R T lnKP

Note: In the reaction, where all gaseous reactants and products; K represents KP, we may conclude that for standard reactions, i.e., at 1 M or 1 atm

When ΔG∘=−ve or K>1: forward reaction is feasible

ΔG∘=+ve or K<1: reverse reaction is feasible

ΔG∘=0 or K=1: reaction is at equilibrium (very rare)

Kc for reaction N2O4⇌2NO2 in chloroform at 291 K is 1.14. Calculate the free energy change of the reaction when the concentration of the two gases are 0.5 mol dm−3 each at the same temperature. (R=0.082 lit atm K−1mol−1)

- −1 lit atm
- −19.67 lit atm
- −5 lit atm
- −13.26 lit atm

**Q.**Calculate ΔS for 3 mol of a diatomic ideal gas which is heated and compressed from 298 K and 1 bar to 398 K and 5 bar. Given: Cp, m=(7/2)R, log(398298)=0.126

- +1.488 J K−1
- −14.883 J K−1
- +14.883 J K−1
- −1.488 J K−1

**Q.**Which of the following is not correct for a gas undergoing isothermal expansion?

- ΔU=0
- ΔT=0
- W=0
- ΔH=0

**Q.**1 mole of an ideal gas at 20 atm pressure and 15 L volume expands such that the final pressure becomes 10 atm and the final volume becomes 60 L. Calculate the change in entropy for the process.

(Cp.m=30 J mole−1 K−1, R=253J/mol K, ln 2 = 0.7, ln 3 = 1.1).

- 80.2 JK−1 mole−1
- 15.17 JK−1 mol−1
- 120×102 JK−1 mol−1
- 26.83 JK−1 mol−1

**Q.**The equilibrium constant for the reaction given below is 2.0×10−7 at 300 K. Calculate the standard entropy change if △H∘=28.40 kJmol−1 for the reaction:

PCl5(g)⇌PCl3(g)+Cl2(g)

- −43.6 Jmol−1K−1
- 43.6 Jmol−1K−1
- −33.6 Jmol−1K−1
- 33.6 Jmol−1K−1

**Q.**When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas (ΔS) is

- Cv.m ln 2
- Cp.m ln 2
- R In 2
- (Cv, m−R) ln 2

**Q.**The equilibrium constant Kc for the gaseous phase reaction at 523 K: PCl3+Cl2⇌PCl5 is 23.10 L mol−1. Calculate Kp at 503 K

- 0.05 atm−1
- 0.65 atm−1
- 0.56 atm−1
- 1.00 atm−1

**Q.**A sample of an ideal gas is expanded to twice its original volume of 1m3 in a reversible process for which P=αv2 where α=5 atm/m6 . If Cv, m=20 Jmol−1K−1, determine the approximate molar change in entropy for the process.

- 4 Jmol−1 K−1
- 47.35 Jmol−1 K−1
- 40 Jmol−1 K−1
- 4 Jmol−1 K−1

**Q.**2 moles of ideal gas at 27oC temperature is expanded reversibly from 2 L to 20 L. Find entropy change: (R= 2 cal/ mol K).

- 0
- 92.1
- 4
- 9.2

**Q.**Find the Entropy change in an isothermal expansion of one mole of an ideal gas from volume V1 to V2.

**Q.**Calculate the boiling point of a solution containing 0.60 g of benzoic acid in 50 g of CS2(l) assuming 90% association of the acid.

The boiling point and Kb for CS2 are 46.2∘C and 2.44 ∘ C kg/mol respectively.

- 47.31 ∘C
- 46.33 ∘C
- 48.94 ∘C
- 46.20 ∘C

**Q.**Standard Free Energy and Equilibrium constant: The change in free energy for a reaction taking place between gaseous reactants and products represented by general equation.

ΔG=ΔG∘+R T lnQP the condition for a system to be at equilibrium is that

ΔG=0 and Qp=KP

Thus at equilibrium

ΔG∘=−R T lnKP

Note: In the reaction, where all gaseous reactants and products; K represents KP, we may conclude that for standard reactions, i.e., at 1 M or 1 atm

When ΔG∘=−ve or K>1: forward reaction is feasible

ΔG∘=+ve or K<1: reverse reaction is feasible

ΔG∘=0 or K=1: reaction is at equilibrium (very rare)

For the equilibrium NiO(s)+CO(g)⇌Ni(s)+CO2(g), ΔG∘(J Mol−1)=−20700−11.97T.. Calculate the temperature at which the product gases at equilibrium at 1 atm will conatin 400 ppm of carbon monoxide.

- 390 K
- 400 K
- 275 K
- 200 K

**Q.**1 mole of an ideal gas at 25∘C is subjected to expand reversibly 10 times of its initial volume. Calculate the change in entropy of expansions.

25.15 J K

^{-1}mol^{-1}29.15 J K

^{-1}mol^{-1}19.15 J K

^{-1}mol^{-1}30.15 J K

^{-1}mol^{-1}

**Q.**Acetic acid dimerises in vapour phase at 437K as 2CH3COOH(g)⇌(CH3COOH)2(g). A mass of 0.0519g of acetic acid was found to have 764.3mm pressure at 437K in a container of 21.45cm3. Calculate degree of dimerisation for acetic acid dimerisation.

**Q.**Which is the correct expression that relates changes of entropy with the change of pressure for an ideal gas at constant temperature, among the following?

- ΔS=nRTlnP2P1
- ΔS=T(P2−P1)
- ΔS=nRln(P1P2)
- ΔS=2.303nRTln(P1P2)

**Q.**When two moles of an ideal gas (Cp, m=52R) is heated from 300 K to 600 K at constant pressure, the change in entropy of the gas (ΔS) is

- 32Rln2
- 3 Rln2
- 5Rln2
- 52Rln2

**Q.**For the water gas reaction:

C(s)+H2O(g)⇌CO(g)+H2(g),

the standard Gibbs energy of reaction (at 1000 K) is −8.1 kJ mol−1. What will be the equilibrium constant for the reaction? log 2.648=0.4229

- 2 mol L−1
- 4.8 mol L−1
- 3.5 mol L−1
- 2.648 mol L−1

**Q.**For the reversible isothermal expansion of one mole of an ideal gas at 300 K, from the volume of 10 dm3 to 20 dm3, ΔH is:

- Zero
- 5.73 kJ
- 1.73 kJ
- 3.46 kJ

**Q.**One mole of an ideal gas at 22.4 L is expanded isothermally and reversibly at 300 K to a volume of 224 L . Which of the following options are correct?

(Given : R=2 cal K−1 mol−1)

- Change in enthalpy is 0
- The magnitude of work done in the process is 1.38 kcal
- Heat change during the process is −1.38 kcal
- Change in internal energy is 10 kcal

**Q.**What is the freezing point of a solution containing 8.1 g HBr in 100gm. water assuming the acid to be 90% ionised:

[Kf for water =1.86Kmolarity−1]

- 0.85∘C
- 0∘C
- −0.35∘C
- −3.53∘C

**Q.**1 mole of an ideal gas at 20 atm pressure and 15 L volume expands such that the final pressure becomes 10 atm and the final volume becomes 60 L. Calculate the change in entropy for the process.

(Cp.m=30 J mole−1 K−1, R=253J/mol K, ln 2 = 0.7, ln 3 = 1.1).

- 80.2 JK−1 mole−1
- 15.17 JK−1 mol−1
- 120×102 JK−1 mol−1
- 26.83 JK−1 mol−1

**Q.**Acetic acid CH3COOH can form a dimer (CH3COOH)2 in the gas phase. The dimer is held together by two H - bonds with a total strength of 77.43 kJ per mol of dimer.

If at 27∘C, the equilibrium constant for the dimerization is 1.096×103, then calculate the magnitude of ΔS∘ (in kJ) for the reaction:

2CH3COOH(g)⇌(CH3COOH)2(g) (Round upto 2 digits after decimal and ln(1.096×103)≈7 and R=8.3 JK−1mol−1)

**Q.**In the reaction:

NH4COONH4(s)⇌2NH3(g)+CO2(g)

Equilibrium pressure for this reaction is 3 atm. The value of Kp will be:

**Q.**When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas (ΔS) is

- Cp.m ln 2
- Cv.m ln 2
- R In 2
- (Cv, m−R) ln 2

**Q.**For the reaction A(g)⇌B(g) at 495 K, ΔG∘= –9.478 kJ mol−1. If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is

[R=8.314 J mol−1K−1; ln10=2.303]

**Q.**When 1 mole of an ideal gas at 20 atm pressure and 15 L volume expands such that the final pressure becomes 10 atm and the final volume becomes 60 L. Calculate entropy change for the process

(Cp, m=30.96Jmole−1K−1)

- 80.2J.k−1mol−1
- 62.42kJ.k−1mol−1
- 120×102Jk−1mol−1
- 27.22J.k−1mol−1

**Q.**One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :

(R=0.082Latmmol−1K−1=8.3mol−1K−1)

- 0
- Rℓn(2490)
- 32Rℓn(24.6)
- Rℓn(24.6)