
A large positive value of $\Delta G$ corresponds to which of these?
(A) Small positive $K$
(B) Small negative $K$
(C) Large positive $K$
(D) Large positive $K$
Answer
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Hint:To solve this question, we must first understand some basic concepts about Electrochemistry and Gibbs Free Energy. Then we need to assess the concept and formulae of Gibbs Free energy to evaluate the effect of Increasing $\Delta G$ and then only we can conclude the correct answer.
Complete step-by-step answer:Before we move forward with the solution of this given question, let us first understand some basic concepts:
Gibbs Free Energy: It is also known as the Gibbs function, Gibbs energy, or free enthalpy, is a quantity that is used to measure the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant. Gibbs free energy is denoted by the symbol ‘G’. Its value is usually expressed in Joules or Kilojoules. Gibbs free energy can be defined as the maximum amount of work that can be extracted from a closed system.
The free energy change of the reaction in any state, ΔG (when equilibrium has not been attained) is related to the standard free energy change of the reaction, ΔG° (which is equal to the difference in the free energies of formation of the products and reactants both in their standard states) according to the equation.
$\Delta G = \,\,\Delta {G^ \circ }\,\, + \,\,RT\ln K$
Where K is the reaction quotient.
At equilibrium, $\Delta G = 0$
So, $\Delta {G^ \circ } = \,\, - RT\ln K$
\[ \Rightarrow K = {e^{ - \Delta {G^ \circ }/RT}} = \dfrac{1}{{{e^{\Delta {G^ \circ }/RT}}}}\,\, > 0\]
Therefore, a large positive value of $\Delta G$ corresponds to small positive K.
So, clearly we can conclude that the correct answer is Option (A).
Note:The Gibbs energy is also the thermodynamic potential that is minimized when a system reaches chemical equilibrium at constant pressure and temperature. Its derivative with respect to the reaction coordinate of the system vanishes at the equilibrium point. As such, a reduction in G is necessary for a reaction to be spontaneous at constant pressure and temperature.
Complete step-by-step answer:Before we move forward with the solution of this given question, let us first understand some basic concepts:
Gibbs Free Energy: It is also known as the Gibbs function, Gibbs energy, or free enthalpy, is a quantity that is used to measure the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant. Gibbs free energy is denoted by the symbol ‘G’. Its value is usually expressed in Joules or Kilojoules. Gibbs free energy can be defined as the maximum amount of work that can be extracted from a closed system.
The free energy change of the reaction in any state, ΔG (when equilibrium has not been attained) is related to the standard free energy change of the reaction, ΔG° (which is equal to the difference in the free energies of formation of the products and reactants both in their standard states) according to the equation.
$\Delta G = \,\,\Delta {G^ \circ }\,\, + \,\,RT\ln K$
Where K is the reaction quotient.
At equilibrium, $\Delta G = 0$
So, $\Delta {G^ \circ } = \,\, - RT\ln K$
\[ \Rightarrow K = {e^{ - \Delta {G^ \circ }/RT}} = \dfrac{1}{{{e^{\Delta {G^ \circ }/RT}}}}\,\, > 0\]
Therefore, a large positive value of $\Delta G$ corresponds to small positive K.
So, clearly we can conclude that the correct answer is Option (A).
Note:The Gibbs energy is also the thermodynamic potential that is minimized when a system reaches chemical equilibrium at constant pressure and temperature. Its derivative with respect to the reaction coordinate of the system vanishes at the equilibrium point. As such, a reduction in G is necessary for a reaction to be spontaneous at constant pressure and temperature.
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