Enthalpies of solution of BaCl 2 ( s ) and BaCl 2 · 2 H 2 O ( s ) are - 20 kJ - 1 mole and 8 . 0 kJ - 1 mole respectively. Calculate heat of hydration of BaCl 2 ( s ) .

Heat of hydration is given by: ∆ H hydration = ∆ H solution - ∆ H lattice Given: ∆ H solution ( BaCl 2 . 2 H 2 O ) = 8 . 0 kJ mol - 1 ∆ H lattice ( BaCl 2 ) = - 20 . 0 kJ mol - 1 Putting the values we get: ∆ H hydration = 8 . 0 - ( - 20 ) = 28 kJ mol - 1

The enthalpy of neutralization of 1 M HCl by 1 M NaOH is - 57 kJ mole - 1 . The enthalpy of formation of water is - 285 kJ - 1 mole . The enthalpy of formation of OH - ion is:

The heat of neutralisation ∆ H neut is the heat set free when one gram equivalent of an acid is neutralized by one gram equivalent of a base in dilute solution at a stated temperature. HCl + NaOH → NaCl + H 2 O , ∆ H rxn = ∆ H neut = - 57 kJ mol - 1 As HCl , NaOH and NaCl are strong acid, strong base and salt, so they ionise 100 % in water. H + + Cl - + Na + + OH - → Na + + Cl - + H 2 O H + + OH - → H 2 O , ∆ H rxn = ∆ H neut = - 57 kJ mol - 1 ......equation 1 We must remember that the heat of formation of H + ion is zero. ∆ H f H + = 0 On applying the heat of formation formula in equation 1 H + + OH - → H 2 O , ∆ H rxn = ∆ H neut = - 57 kJ mol - 1 ∆ H rxn = ∆ H f H 2 O - ∆ H f H + - ∆ H f OH - - 57 = - 285 - 0 - ∆ H f OH - ∆ H f OH - = - 228 kJ mol - 1

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.

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The standard enthalpy of decomposition of the yellow complex $H_{3} {NSO}_{2}$ into ${NH}_{3}$ and ${SO}_{2}$ is $+ 40 kJ {mol}^{- 1} .$ Calculate the standard enthalpy of the formation of $H_{3} {NSO}_{2} . ({ΔH}_{f}^{0} ({NH}_{3}) = - 46.17 kJ {mol}^{- 1}, {ΔH}^{0} {}_{f}{({SO}_{2})} = - 296.83 kJ {mol}^{- 1})$

Important Questions on Thermodynamics

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 134

Calculate the bond energy of $Cl - Cl$ bond from the following data:

${CH}_{4} (g) + {Cl}_{2} (g) ⟶ {CH}_{3} Cl (g) + HCl (g); ΔH = - 100.3 kJ .$ Also the bond enthalpies of $C - H, C - Cl, H - Cl$ bonds are $413, 326$ and $431 kJ {mol}^{- 1}$ respectively.

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 137

Calculate change in enthalpy for the reaction $H_{2} (g) + {Cl}_{2} (g) ⟶ 2 H - Cl (g)$ at $27 ° C$ by using the bond energy and energy data

Bond energies of $H - H, Cl - Cl$ and $H - Cl$ bonds are $435 kJ {mol}^{- 1}, 240 kJ {mol}^{- 1}$ and $430 kJ {mol}^{- 1}$
respectively.

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 140

Enthalpies of solution of

{BaCl}_{2} (s)

and

{BaCl}_{2} \cdot 2 H_{2} O (s)

are

- 20 {kJ}^{- 1} mole

and

8.0 {kJ}^{- 1} mole

respectively.
Calculate heat of hydration of

{BaCl}_{2} (s) .

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 141

Setup of Born-Haber cycle; calculate lattice energy of

{MgO}_{(s)} .

The given that enthalpy of formation of

{MgO}_{(s)} = - 602,

sublimation of

{Mg}_{(s)} = 148; 1^{st}

and

2^{nd}

ionization energy of

Mg = 738

and

1450

respectively. For Oxygen bond dissociation energy

= 498; 1^{st}

2^{nd}

and electron gain enthalpy

= - 141

and

844

respectively (all unit in

kJ {mole}^{- 1}

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 143

The enthalpy of neutralization of

1 M HCl

1 M NaOH

- 57 kJ {mole}^{- 1} .

The enthalpy of formation of water is

- 285 {kJ}^{- 1} mole .

The enthalpy of formation of

{OH}^{-}

ion is:

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 145

The free energy change for a reversible reaction at equilibrium is:

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 146

ΔH °

for the reaction

X_{(g)} + Y_{(g)} ⇌ Z_{(g)}

- 4.6 Kcal,

the value of

ΔU °

227 ° C

(R = 2 cal . {mol}^{- 1} K^{- 1}) :

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-1>Q 148

Consider the reaction at $300 K$

$H_{2} (g) + {Cl}_{2} (g) ⟶ 2 HCl (g); △ H ° = - 185 kJ$

If $2$ moles of $H_{2}$ completely react with $2$ mole of ${Cl}_{2}$ to form $HCl$ , what is $△ U °$ for this reaction.

The standard enthalpy of decomposition of the yellow complex H3NSO2 into NH3 and SO2 is +40 kJ mol-1. Calculate the standard enthalpy of the formation of H3NSO2. (ΔHf0NH3=-46.17 kJ mol-1, ΔH0SO2f=-296.83 kJ mol-1)

Important Questions on Thermodynamics

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