Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution
3. Answer in brief.
i. Obtain the relationship between ΔG0 of a reaction and the equilibrium constant
Answer:-
The Gibbs energy change for a chemical reaction is expressed as:
\[
\Delta G = \Delta G^\circ + RT \ln Q \quad \text{(1)}
\]
Where:
- \(\Delta G^\circ\) represents the standard Gibbs energy change, which is the Gibbs energy change when the reactants and products in a reaction are in their standard states.
- \(Q\) is called the reaction quotient and is analogous to the equilibrium constant. It involves nonequilibrium concentrations or partial pressures in the case of a gaseous reaction.
Consider the following reaction:
\[
aA + bB \rightarrow cC + dD
\]
From equation (1), we have:
\[
\Delta G = \Delta G^\circ + RT \ln Q_C \quad \text{or} \quad \Delta G = \Delta G^\circ + RT \ln Q_P
\]
This can be further expressed as:
\[
\Delta G = \Delta G^\circ + RT \ln\left(\frac{[C]^c[D]^d}{[A]^a[B]^b}\right) \quad \text{or} \quad \Delta G = \Delta G^\circ + RT \ln\left(\frac{P_C^c \cdot P_D^d}{P_A^a \cdot P_B^b}\right)
\]
When the reaction reaches equilibrium,
Δ
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∘
=
0
ΔG
∘
=0, and
�
�
Q
C
and
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Q
P
become
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K
C
and
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K
P
, respectively.
\[
0 = \Delta G^\circ + RT \ln K_C \quad \text{or} \quad 0 = \Delta G^\circ + RT \ln K_P
\]
Which can be simplified to:
\[
\Delta G^\circ = -RT \ln K_C \quad \text{or} \quad \Delta G^\circ = -RT \ln K_P
\]
And in a more commonly used form:
\[
\Delta G^\circ = -2.303RT \log_{10} K_C \quad \text{or} \quad \Delta G^\circ = -2.303RT \log_{10} K_P
\]
Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution