
How does delta G affect Keq?
Answer
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Hint: The answer to this question is dependent on the concept of the spontaneity of the reaction where the change in the spontaneity with different values of the Gibb’s energy will give you the required answer.
Complete step by step answer:
In the lower classes of chemistry, we have dealt with the reaction dynamics chapter of the physical chemistry part where we have come across many parameters like types of the reaction, enthalpy of the reaction, entropy of a reaction and also about the Gibbs free energy change of the reaction.
- Let us now see in detail about how the equilibrium constant ${{K}_{eq}}$ affects the Gibbs free energy that is $\Delta G$.
- Gibbs free energy is defined as the energy associated with a chemical reaction which can be used to do the work. It is the sum of enthalpy and the product of temperature and entropy of the system.
- This Gibbs free energy relates to the equilibrium constant in direct ways and there are three possibilities and they are,
1) If the value of Gibb’s energy is zero, that is if $\Delta G = 0$, then the reaction is at equilibrium and there is no net change taking place.
2) If $\Delta G=-ve$, then the reaction is spontaneous and thus proceeds in the forward reaction.
3) If $\Delta G=+ve$, then the reaction is nonspontaneous in the forward direction and the reverse reaction will be spontaneous and proceeds towards the reactant side.
The mathematical expression can be shown as,
$\Delta G = \Delta {{G}^{0}}+RT\ln Q$
At equilibrium, $\Delta G=0$and Q = K
\[\Rightarrow 0 = \Delta {{G}^{0}}+RT\ln K\]
\[\Rightarrow \ln K = \dfrac{\Delta {{G}^{0}}}{RT}\]
\[\Rightarrow K = {{e}^{-\dfrac{\Delta {{G}^{0}}}{RT}}}\]
Thus, if $\Delta G = -ve$, then\[-\dfrac{\Delta {{G}^{0}}}{RT}\] is positive and\[{{e}^{-\dfrac{\Delta {{G}^{0}}}{RT}}}\]> 1 which means that K > 1 and the reaction is spontaneous in forward direction.
Similarly, if $\Delta G = +ve$, then K < 1 and this suggests that the forward reaction is nonspontaneous.
Note: Note that the spontaneity of the reaction is the one that occurs without the addition of external energy and this process may take place quickly or slowly because spontaneity is not related to kinetics or the reaction rate.
Complete step by step answer:
In the lower classes of chemistry, we have dealt with the reaction dynamics chapter of the physical chemistry part where we have come across many parameters like types of the reaction, enthalpy of the reaction, entropy of a reaction and also about the Gibbs free energy change of the reaction.
- Let us now see in detail about how the equilibrium constant ${{K}_{eq}}$ affects the Gibbs free energy that is $\Delta G$.
- Gibbs free energy is defined as the energy associated with a chemical reaction which can be used to do the work. It is the sum of enthalpy and the product of temperature and entropy of the system.
- This Gibbs free energy relates to the equilibrium constant in direct ways and there are three possibilities and they are,
1) If the value of Gibb’s energy is zero, that is if $\Delta G = 0$, then the reaction is at equilibrium and there is no net change taking place.
2) If $\Delta G=-ve$, then the reaction is spontaneous and thus proceeds in the forward reaction.
3) If $\Delta G=+ve$, then the reaction is nonspontaneous in the forward direction and the reverse reaction will be spontaneous and proceeds towards the reactant side.
The mathematical expression can be shown as,
$\Delta G = \Delta {{G}^{0}}+RT\ln Q$
At equilibrium, $\Delta G=0$and Q = K
\[\Rightarrow 0 = \Delta {{G}^{0}}+RT\ln K\]
\[\Rightarrow \ln K = \dfrac{\Delta {{G}^{0}}}{RT}\]
\[\Rightarrow K = {{e}^{-\dfrac{\Delta {{G}^{0}}}{RT}}}\]
Thus, if $\Delta G = -ve$, then\[-\dfrac{\Delta {{G}^{0}}}{RT}\] is positive and\[{{e}^{-\dfrac{\Delta {{G}^{0}}}{RT}}}\]> 1 which means that K > 1 and the reaction is spontaneous in forward direction.
Similarly, if $\Delta G = +ve$, then K < 1 and this suggests that the forward reaction is nonspontaneous.
Note: Note that the spontaneity of the reaction is the one that occurs without the addition of external energy and this process may take place quickly or slowly because spontaneity is not related to kinetics or the reaction rate.
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