Chemistry Class 11 NCERT Solutions: Chapter 6 Thermodynamics Part 4

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Q: 15 Calculate the enthalpy change for the process

and calculate bond enthalpy of

is enthalpy of atomization

Answer:

The chemical equations implying to the given values of enthalpies are:

(i)

(iii)

(iv)

Enthalpy change for the given process can be calculated using the following calculations as:

Equation (ii) +2× Equation (iii) – Equation (i) – Equation (iv)

Bond enthalpy of C-Cl bond in

Q: 16. For an isolated system, , what will be ?

Answer:

will be positive i.e., greater than zero

Since will be positive and the reaction will be spontaneous.

Q: 17. For the reaction at ,

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range?

Answer:

From the expression,

Assuming the reaction at equilibrium, ∆T for the reaction would be:

For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

Q: 18. For the reaction,

Answer:

The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore energy is being released. Hence, ∆H is negative.

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ∆S is negative for the given reaction.

Q: 19. For the reaction

Calculate for the reaction, and predict whether the reaction spontaneously.

Answer: For the given reaction,

Substituting the value of in the expression of

Substituting the value of in the expression of

Since for the reaction is positive, the reaction will not occur spontaneously.