Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>Competition Focus>Q 101

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IMPORTANT

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Reactions that have standard free energy change less than zero always have equilibrium constant to be

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Important Questions on Thermodynamics

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IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>Competition Focus>Q 102

Standard entropies of

X_{2}, Y_{2}

and

{XY}_{3}

are

60, 40 and 50 J K^{- 1} {mol}^{- 1}

respectively. For the reaction

\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ {XY}_{3}, ∆ H = - 30 KJ

to be at equilibrium, the temperature should be

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Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>Competition Focus>Q 103

For the reaction

X_{2} O_{2} (l) \to 2 {XO}_{2} (g), ∆ U = 2.1 KCal

,

∆ S = 20 cal K^{- 1}

at

300 K

. Hence,

∆ G

is

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Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>Competition Focus>Q 104

Some substances possess some residual entropy even at absolute zero. This is because

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Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>Competition Focus>Q 105

When one mole of a monatomic ideal gas at initial temperature

T K

expands adiabatically from

1

litre to

2

litre, the final temperature in Kelvin would be

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Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>Competition Focus>Q 106

When

1 mol

of

{CO}_{2} (g)

occupying a volume of

10 L at 27^{\circ} C

is allowed to expand under adiabatic conditions, temperature falls to

- 123^{\circ} C

. Hence, final volume of the gas will be:

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Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>Competition Focus>Q 107

1 mol of {NH}_{3}

gas

(γ = 1.33)

at

27^{\circ} C

is allowed to expand adiabatically so that the final volume becomes

8

times. Work done will be

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Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>Competition Focus>Q 108

Temperature of

1

mole of a gas is increased by

1^{\circ} C

at constant pressure. Work done by the system is

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Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>Competition Focus>Q 109

For the reaction,

H_{2} (g) + \frac{1}{2} O_{2} (g) ➔ H_{2} O (l)

,

∆ C_{p} = 32 {JK}^{- 1}

,

∆ H

at

27^{\circ} C = - 285 KJ {mol}^{- 1} .

What will be the value of

∆ H

at

127^{\circ} C

?