State the second law of thermodynamics.

The second law of thermodynamics can be expressed in terms of entropy. It can be stated as “the entropy of an isolated system increases during a spontaneous process”. The total entropy change ( ∆ S total ) for the system and the surroundings of a spontaneous process is given by, ∆ S total = ∆ S system + ∆ S surrounding > 0 When a system is in equilibrium, the entropy is maximum and the change in entropy, ∆ S = 0 . Hence, entropy of a spontaneous process increases till it reaches a maximum and at equilibrium the change in entropy is zero.

The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at 25 ° C is

298 × 10 7 × 8 . 31 × 2 . 303 log 2 .

2 . 303 × 298 × 0 . 082 log 2 .

2 . 303 × 298 × 0 . 082 log 0 . 5 .

8 . 31 × 10 7 × 298 × 2 . 303 log 0 . 5 .

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by:

with increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases.

with increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive.

with increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative.

E M B I B E

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 151

EASY

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IMPORTANT

Earn 100

Calculate the Standard internal energy of formation of liquid methyl acetate $({CH}_{3} {COOCH}_{3})$ from its standard enthalpy of formation, which is $- 442.91 kJ {mole}^{- 1}$ at $25 ° C .$

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Important Questions on Thermodynamics

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 152

2 C + O_{2} \to 2 C O; Δ H = - 220 k J

Which of the following statement is correct for this reaction

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 153

$C (s) + O_{2} (g) ⟶ {CO}_{2}, (g);$ $ΔH = - 94.3 {kcal}^{- 1} mol$

$CO (g) + \frac{1}{2} O_{2} (g) ⟶ {CO}_{2} (g);$ $ΔH = - 67.4 {kcal}^{- 1} mol$

$O_{2} (g) ⟶ 2 O (g);$ $ΔH = 117.4 {kcal}^{- 1} mol$

$CO (g) ⟶ C (g) + O (g);$ $ΔH = 230.6 {kcal}^{- 1} mol$

Calculate $ΔH$ for $C (s) ⟶ C (g)$ in ${kcal}^{- 1} mol .$

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 154

In the reaction

{CO}_{2} (g) + H_{2} (g) ⟶ CO (g) + H_{2} O (g)

ΔH = 40 kJ; ΔH

represents :

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 155

Given

H_{2} (g) + {Br}_{2} (g) \to 2 HBr (g), {ΔH}_{1} °

and standard enthalpy of condensation of bromine is

{ΔH}_{2} °

, standard enthalpy of formation of

HBr

at

25 ° C

is:

EASY

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 156

For the following reaction,

$C$ (diamond) $+ O_{2} ⟶ {CO}_{2} (g); ΔH = - 94.3 kcal {mol}^{- 1}$

$C$ (graphite) $+ O_{2} ⟶ {CO}_{2} (g); ΔH = - 97.6 kcal {mol}^{- 1}$

The heat required to change $1 g$ of $C$ (graphite) $⟶$ (diamond) is

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 157

The standard heat of combustion of solid boron is equal to:

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 158

The heat of combustion of sucrose

(C_{12} H_{22} O_{11})

is

1350 K cal

. How much of heat will be liberated when

17.1 g

of sucrose is burnt?

EASY

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 159

If $S + O_{2} ⟶ {SO}_{2}, ΔH = - 298.2 kJ {mole}^{- 1}$

${SO}_{2} + 1 / 2 O_{2} ⟶ {SO}_{3}, ΔH = - 98.7 kJ {mole}^{- 1}$

${SO}_{3} + H_{2} O ⟶ H_{2} {SO}_{4}, ΔH = - 130.2 kJ {mole}^{- 1}$

$H_{2} + 1 / 2 O_{2} ⟶ H_{2} O, ΔH = - 287.3 kJ {mole}^{- 1}$

The enthalpy of formation of $H_{2} {SO}_{4}$ at $298 K$ will be:

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Calculate the Standard internal energy of formation of liquid methyl acetate CH3COOCH3 from its standard enthalpy of formation, which is -442.91 kJ mole-1at 25°C.

Important Questions on Thermodynamics

Learn and Prepare for any exam you want !

Calculate the Standard internal energy of formation of liquid methyl acetate $({CH}_{3} {COOCH}_{3})$ from its standard enthalpy of formation, which is $- 442.91 kJ {mole}^{- 1}$ at $25 ° C .$