Ncert Solutions Chemistry Class 11th

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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.

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Heat has a randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. The mathematical relation which relates these three parameters is? S = q_rev/ T

Here? S = change in entropy  ^

q_rcv = heat of reversible reaction '

T = temperature

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It is a spontaneous process. Although enthalpy change is zero, randomness or disorder (ΔS) increases and ΔS is positive. Therefore, in the equation, ΔG = ΔH – TΔS, the term TΔS will be negative. Hence ΔG will be negative.