The free energy changes for the two reactions given below are: SO 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 2270 cal . S ( rhom ) + O 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 74060 cal . Find ∆ G for the reaction: S ( rhom ) + O 2 ( g ) → SO 2 ( g ) .

Given, SO 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 2270 cal ----- i S ( rhom ) + O 2 ( g ) + Cl 2 ( g ) → SO 2 Cl 2 ( g ) , ∆ G = - 74060 cal ----- ii The free energy changes for two reactions are given above. To calculate Gibbs free energy, Eq ii - Eq i , = ∆ G ii - ∆ G i . = - 74060 - - 2270 cal . = - 71790 cal . Here, the free energy change value is negative. i.e. it is also a spontaneous process.

At what temperature, reduction of lead oxide to lead by carbon, i.e., PbO ( s ) + C ( s ) → Pb ( s ) + CO ( g ) becomes spontaneous. For the reaction, ∆ H and ∆ S are 108 . 4 kJ mol - 1 and 190 . 0 J K - 1 mol - 1 respectively.

Given, PbO ( s ) + C ( s ) → Pb ( s ) + CO ( g ) . ∆ H = 108 . 4 kJ mol - 1 , ∆ S = 190 J K - 1 mol - 1 . At equilibrium, the free energy change, ∆ G = 0 . Formula Used, ∆ G = ∆ H - T ∆ S . 0 = ∆ H - T ∆ S . ∆ H = T ∆ S . T = ∆ H ∆ S . T = 108400 J mol - 1 190 J K - 1 mol - 1 = 570 . 5 K . (Above 570 .5 K , ∆ G , the free energy will be negative and reaction will be spontaneous).

For the reaction: 2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g ) , the enthalpy and entropy changes are - 113 · 0 kJ mol - 1 and - 145 J K - 1 mol - 1 respectively. Find the temperature below which this reaction is spontaneous.

Given, 2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g ) . ∆ H = - 113 . 0 kJ mol - 1 , ∆ S = - 145 J K - 1 mol - 1 . At equilibrium, ∆ G = 0 . Formula Used, ∆ G = ∆ H - T ∆ S . T = ∆ H ∆ S . T = - 113 × 1000 - 145 T = 779 .3 K . (Below 779 .3 K , ∆ G will be negative and reaction will be spontaneous).

E M B I B E

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 3

MEDIUM

11th CBSE

IMPORTANT

Earn 100

$∆ H$ and $∆ S$ for the vaporization of water at $1 atm$ pressure are $40.63 kJ {mol}^{- 1}$ and $108.8 J K^{- 1} {mol}^{- 1}$ respectively. Calculate the temperature at which the free energy for this transformation will be zero. Predict, giving reasons, the sign of free energy change $(∆ G)$ above this temperature.

Important Questions on Thermodynamics

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 4

In the reaction $A^{+} + B \to A + B^{+}$ , there is no entropy change. Enthalpy change, $∆ H$ , equals $22 kJ {mol}^{- 1}$ of $A^{+}$ . Calculate $∆ G$ for the reaction. Under what condition free energy change will have negative value?

EASY

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 5

Calculate the free energy change in dissolving one mole of sodium chloride at

25 ° C

. Lattice energy

= + 777.8 kJ {mol}^{- 1}

. Hydration energy of

NaCl = - 774.1 kJ {mol}^{- 1}

and

∆ S

at

25 ° C = 0.043 kJ K^{- 1} {mol}^{- 1}

.

MEDIUM

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 6

The following data is known about the melting of

KCl

:

∆ H = 7.25 kJ {mol}^{- 1}

and

∆ S = 0.007 kJ K^{- 1} {mol}^{- 1}

. Calculate the melting point.

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 7

The free energy changes for the two reactions given below are:

${SO}_{2} (g) + {Cl}_{2} (g) \to {SO}_{2} {Cl}_{2} (g), ∆ G = - 2270 cal$ .

$S (rhom) + O_{2} (g) + {Cl}_{2} (g) \to {SO}_{2} {Cl}_{2} (g), ∆ G = - 74060 cal$ .

Find $∆ G$ for the reaction: $S (rhom) + O_{2} (g) \to {SO}_{2} (g)$ .

HARD

11th CBSE

IMPORTANT

Pradeep's Chemistry Vol 1>Chapter 7 - Thermodynamics>PROBLEMS FOR PRACTICE>Q 8

At what temperature, reduction of lead oxide to lead by carbon, i.e., $PbO (s) + C (s) \to Pb (s) + CO (g)$ becomes spontaneous. For the reaction, $∆ H$ and $∆ S$ are $108.4 kJ {mol}^{- 1}$ and $190.0 J K^{- 1} {mol}^{- 1}$ respectively.