Entropy for a...

Entropy for a Reversible Process

Trending Questions

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

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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

25^{o} C,

the equilibrium constant for the dimerisation is

1.3 \times 10^{3}

(pressure in atm). What is magnitude

△ S^{0}

for the reaction? Assume that

△ H

does not vary with temperature.

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Q. 1 mole of an ideal gas is allowed to expand isothermally at

27^{o} C

until its volume is tripled. Calculate

△ S_{s y s}

and

△ S_{u n i v}

under the reversible expansion.

$△ S_{s y s} = 9.1 J K^{- 1} m o l^{- 1}$
$△ S_{u n i v} = △ S_{s y s} = 9.1 J K^{- 1} m o l^{- 1}$
$△ S_{s u r r} = - 9.1 J K^{- 1} m o l^{- 1}$
$△ S_{u n i v} = 0$

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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.

Δ G = Δ G^{\circ} + R T l n Q_{P}

the condition for a system to be at equilibrium is that

Δ G = 0 and Q_{p} = K_{P}

Thus at equilibrium

Δ G^{\circ} = - R T l n K_{P}

Note: In the reaction, where all gaseous reactants and products;

K represents K_{P}

, we may conclude that for standard reactions, i.e., at

1 M or 1 atm

When

Δ G^{\circ} = - v e or K > 1 : forward reaction is feasible

Δ G^{\circ} = + v e or K < 1 : reverse reaction is feasible

Δ G^{\circ} = 0 or K = 1 : reaction is at equilibrium (very rare)

K_{c}

for reaction

N_{2} O_{4} ⇌ 2 N O_{2}

in chloroform at

291 K

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}^{- 1} {mol}^{- 1}

)

$- 1 lit atm$
$- 19.67 lit atm$
$- 5 lit atm$
$- 13.26 lit atm$

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Q. Calculate

Δ S

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:

C_{p, m} = (7 / 2) R

log (\frac{398}{298}) = 0.126

$+ 1.488 {J K}^{- 1}$
$- 14.883 {J K}^{- 1}$
$+ 14.883 {J K}^{- 1}$
$- 1.488 {J K}^{- 1}$

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

$Δ U = 0$
$Δ T = 0$
$W = 0$
$Δ H = 0$

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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.

(C_{p . m} = 30 {J mole}^{- 1} K^{- 1}, R = \frac{25}{3} J/mol K, ln 2 = 0.7, ln 3 = 1.1)

$80.2 {JK}^{- 1} {mole}^{- 1}$
$15.17 {JK}^{- 1} {mol}^{- 1}$
$120 \times 10^{2} {JK}^{- 1} {mol}^{- 1}$
$26.83 {JK}^{- 1} {mol}^{- 1}$

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Q. The equilibrium constant for the reaction given below is

2.0 \times 10^{- 7}

at 300 K. Calculate the standard entropy change if

△ H^{\circ} = 28.40 k J m o l^{- 1}

for the reaction:

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

$- 43.6 J m o l^{- 1} K^{- 1}$
$43.6 J m o l^{- 1} K^{- 1}$
$- 33.6 J m o l^{- 1} K^{- 1}$
$33.6 J m o l^{- 1} K^{- 1}$

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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

(Δ S)

$C_{v . m}$ ln 2
$C_{p . m}$ ln 2
R In 2
$(C_{v, m} - R)$ ln 2

View Solution

Q. The equilibrium constant

K_{c}

for the gaseous phase reaction at

523 K

P C l_{3} + C l_{2} ⇌ P C l_{5}

23.10 {L mol}^{- 1}

. Calculate

K_{p}

503 K

$0.05 {atm}^{- 1}$
$0.65 {atm}^{- 1}$
$0.56 {atm}^{- 1}$
$1.00 {atm}^{- 1}$

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Q. A sample of an ideal gas is expanded to twice its original volume of

1 m^{3}

in a reversible process for which

P = α v^{2}

where α=5

a t m / m^{6}

. If

C_{v, m} = 20 J m o l^{- 1} K^{- 1}

, determine the approximate molar change in entropy for the process.

$4 J m o l^{- 1} K^{- 1}$
$47.35 J m o l^{- 1} K^{- 1}$
$40 J m o l^{- 1} K^{- 1}$
$4 J m o l^{- 1} K^{- 1}$

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Q. 2 moles of ideal gas at

27^{o} C

temperature is expanded reversibly from 2 L to 20 L. Find entropy change: (R= 2 cal/ mol K).

0
92.1
4
9.2

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

V_{1}

V_{2}

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Q. Calculate the boiling point of a solution containing 0.60 g of benzoic acid in 50 g of

C S_{2} (l)

assuming 90% association of the acid.
The boiling point and

K_{b}

for

C S_{2}

are

{46.2}^{\circ} C

and

2.44^{\circ} C kg/mol

respectively.

$47.31^{\circ} C$
$46.33^{\circ} C$
$48.94^{\circ} C$
$46.20^{\circ} C$

View Solution

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.

Δ G = Δ G^{\circ} + R T l n Q_{P}

the condition for a system to be at equilibrium is that

Δ G = 0 and Q_{p} = K_{P}

Thus at equilibrium

Δ G^{\circ} = - R T l n K_{P}

Note: In the reaction, where all gaseous reactants and products;

K represents K_{P}

, we may conclude that for standard reactions, i.e., at

1 M or 1 atm

When

Δ G^{\circ} = - v e or K > 1 : forward reaction is feasible

Δ G^{\circ} = + v e or K < 1 : reverse reaction is feasible

Δ G^{\circ} = 0 or K = 1 : reaction is at equilibrium (very rare)

For the equilibrium

N i O (s) + C O (g) ⇌ N i (s) + C O_{2} (g), Δ G^{\circ} ({J Mol}^{- 1}) = - 20700 - 11.97 T .

. Calculate the temperature at which the product gases at equilibrium at

1 atm

will conatin

400 ppm

of carbon monoxide.

$390 K$
$400 K$
$275 K$
$200 K$

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Q. 1 mole of an ideal gas at

25^{\circ} C

is subjected to expand reversibly 10 times of its initial volume. Calculate the change in entropy of expansions.

25.15 J K^-1mol^-1
29.15 J K^-1mol^-1
19.15 J K^-1mol^-1
30.15 J K^-1mol^-1

View Solution

Q. Acetic acid dimerises in vapour phase at

437 K

2 C H_{3} C O O H (g) ⇌ {(C H_{3} C O O H)}_{2} (g)

. A mass of

0.0519 g

of acetic acid was found to have

764.3 m m

pressure at

437 K

in a container of

21.45 {c m}^{3}

. Calculate degree of dimerisation for acetic acid dimerisation.

View Solution

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?

$Δ S = n R T ln \frac{P_{2}}{P_{1}}$
$Δ S = T (P_{2} - P_{1})$
$Δ S = n R ln (\frac{P_{1}}{P_{2}})$
$Δ S = 2.303 n R T ln (\frac{P_{1}}{P_{2}})$

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Q. When two moles of an ideal gas

(C_{p, m} = \frac{5}{2} R)

is heated from

300

K to

600

K at constant pressure, the change in entropy of the gas (

Δ

S) is

$\frac{3}{2} R l n 2$
$3 R l n 2$
$5 R l n 2$
$\frac{5}{2} R l n 2$

View Solution

Q. For the water gas reaction:

C_{(s)} + H_{2} O_{(g)} ⇌ C O_{(g)} + H_{2 (g)},

the standard Gibbs energy of reaction (at 1000 K) is

- 8.1 k J m o l^{- 1} .

What will be the equilibrium constant for the reaction?

l o g 2.648 = 0.4229

$2 m o l L^{- 1}$
$4.8 m o l L^{- 1}$
$3.5 m o l L^{- 1}$
$2.648 m o l L^{- 1}$

View Solution

Q. For the reversible isothermal expansion of one mole of an ideal gas at 300 K, from the volume of

10 {d m}^{3}

20 {d m}^{3}, Δ H

is:

Zero
$5.73 k J$
$1.73 k J$
$3.46 k J$

View Solution

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 $10 kcal$

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Q. What is the freezing point of a solution containing

8.1 g HBr

100 g m

. water assuming the acid to be

90 %

ionised:

[

K_{f}

for water

= 1.86 K {molarity}^{- 1}

]

${0.85}^{\circ} C$
$0^{\circ} C$
$- {0.35}^{\circ} C$
$- {3.53}^{\circ} C$

View Solution

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.

(C_{p . m} = 30 {J mole}^{- 1} K^{- 1}, R = \frac{25}{3} J/mol K, ln 2 = 0.7, ln 3 = 1.1)

$80.2 {JK}^{- 1} {mole}^{- 1}$
$15.17 {JK}^{- 1} {mol}^{- 1}$
$120 \times 10^{2} {JK}^{- 1} {mol}^{- 1}$
$26.83 {JK}^{- 1} {mol}^{- 1}$

View Solution

Q. Acetic acid

C H_{3} C O O H

can form a dimer

(C H_{3} C O O H)_{2}

in the gas phase. The dimer is held together by two H - bonds with a total strength of

77.43 k J p e r m o l

of dimer.

If at

27^{\circ} C

, the equilibrium constant for the dimerization is

1.096 \times 10^{3}

, then calculate the magnitude of

Δ S^{\circ}

(in

k J

) for the reaction:

2 C H_{3} C O O H (g) ⇌ (C H_{3} C O O H)_{2} (g)

(Round upto 2 digits after decimal and

ln (1.096 \times 10^{3}) \approx 7 a n d R = 8.3 J K^{- 1} m o l^{- 1}

)

View Solution

Q. In the reaction:

{N H}_{4} C O O {N H}_{4} (s) ⇌ 2 {N H}_{3} (g) + {C O}_{2} (g)

Equilibrium pressure for this reaction is

3

atm. The value of

K_{p}

will be:

View Solution

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

(Δ S)

$C_{p . m}$ ln 2
$C_{v . m}$ ln 2
R In 2
$(C_{v, m} - R)$ ln 2

View Solution

Q. For the reaction

A (g) ⇌ B (g)

495 K, Δ G^{\circ} = - 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 millimoles. (Round off to the nearest Integer).

[R = 8.314 {J mol}^{- 1} K^{- 1}; l n 10 = 2.303]

View Solution

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

(C_{p, m} = 30.96 J m o l e^{- 1} K^{- 1})

$80.2 J . k^{- 1} m o l^{- 1}$
$62.42 k J . k^{- 1} m o l^{- 1}$
$120 \times 10^{2} J k^{- 1} m o l^{- 1}$
$27.22 J . k^{- 1} m o l^{- 1}$

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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 = 0.082 L a t m {m o l}^{- 1} K^{- 1} = 8.3 {m o l}^{- 1} K^{- 1}

)

0
$R ℓ n (2490)$
$\frac{3}{2} R ℓ n (24.6)$
$R ℓ n (24.6)$

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