Chemistry NCERT Exemplar Solutions Class 11th Chapter Six

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At constant volume

q = ΔU + (-w)

-w = pΔ q = AU + pΔV

ΔV = 0 (at constant volume)

Hence, qv = ΔU + 0 = ΔU= change in internal energy at constant pressure, qp = AU + pΔV

Since ΔU + pΔV=ΔH

=> qp = ΔH change in enthalpy

Hence, at constant volume and at constant pressure, heat change is a state function because it is equal to ΔU and ΔH respectively which are state functions.

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For an isolated system w = 0, q = 0

Since ΔU= q + w = 0 + 0 = 0, ΔU= 0

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Gibbs energy for a reaction in which all reactants and products are in standard state. ΔrG° is related to the equilibrium constant of the reaction as follows

ΔrG = ArG° + RT In K

At equilibrium, 0 = ΔrG° + RT InA– ( {ΔrG = 0) or  ΔrG° =-RT lnK

ΔrG° = 0 when K= 1

For all other values of K, ArG° will be non-zero.

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Molar enthalpy of vaporization is more for water due to hydrogen bonding between water molecules.

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State functions: Enthalpy, entropy, temperature, free energy Path functions: Heat, work.

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The standard molar entropy of H₂0 (1) is 70 J K^-1 mol^-1. The solid form of H₂0 is ice. In ice, molecules of H₂0 are less random than in liquid water. Thus, molar entropy of H₂0 (s) < molar entropy of H₂0 (1). The standard molar entropy of H₂0 (s) is less than 70 J K^-1 mol^-1.

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During cyclic process, change in internal energy is zero.

ΔU = 0

and no work is said to be done, as system returns to the initial state.
For a steady state cyclic process at any given stage enthalpy is one single value however, at different stage it would vary.

ΔH = 0

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ΔrG° = -RT ln Kp

= -RT ln (0.98)

Since In (0.98) is negative

.'. ΔrG° is positive the reaction is non spontaneous

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Yes, when the system and the surroundings are in thermal equilibrium, their temperatures are the same.