Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

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 � ∘ = 0 ΔG ∘ =0, and � � Q C ​ and � � Q P ​ become � � K C ​ and � � 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: \[ \Delta G^\circ = -2.303RT \log_{10} K_C \quad \text{or} \quad \Delta G^\circ = -2.303RT \log_{10} K_P \]