
The Gibbs energy change for a reversible reaction at equilibrium is
A.Zero
B.Small positive
C.Small negative
D.Large positive
Answer
584.1k+ views
Hint: Gibbs Free Energy is the energy associated with a chemical reaction that can be used to do work. Free energy of the reaction at equilibrium can be calculated as the sum of free energies of the reactant minus the sum of free energies of the product.
Complete answer:
As we know that in a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentration that has no further tendency to change with time, so that there is no observable change in the properties of the system.
Therefore, for a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy and temperature thereby eliminating $\Delta \text{H}$ from the equation for$\Delta \text{G}$. The general relationship can be shown as follow
$\Delta \text{G=V}\Delta \text{P-S}\Delta \text{T}$
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lower cases letter correspond to the stoichiometric coefficients for the various species:
$\text{aA+bB}\to \text{cC+dD}$
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can derive the following expression for $\Delta \text{G}$-
$\Delta \text{G=}\sum\limits_{\text{m}}{{{\text{G}}_{\text{product}}}}-\sum\limits_{\text{n}}{{{\text{G}}_{\text{reactant}}}}$
Also, Gibbs free energy is constant at equilibrium. Therefore, the sum of the free energies of the products and the reactants is same at equilibrium
Hence, Gibbs energy change is zero for a reversible reaction at equilibrium is option (a). zero.
Note: We should know that the total free energy of the system (reactants + products) always becomes more negative as the reaction proceeds. Eventually it reaches a minimum value at a system composition that defines the equilibrium composition of the system, after which time no further net change will occur.
Complete answer:
As we know that in a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentration that has no further tendency to change with time, so that there is no observable change in the properties of the system.
Therefore, for a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy and temperature thereby eliminating $\Delta \text{H}$ from the equation for$\Delta \text{G}$. The general relationship can be shown as follow
$\Delta \text{G=V}\Delta \text{P-S}\Delta \text{T}$
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lower cases letter correspond to the stoichiometric coefficients for the various species:
$\text{aA+bB}\to \text{cC+dD}$
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can derive the following expression for $\Delta \text{G}$-
$\Delta \text{G=}\sum\limits_{\text{m}}{{{\text{G}}_{\text{product}}}}-\sum\limits_{\text{n}}{{{\text{G}}_{\text{reactant}}}}$
Also, Gibbs free energy is constant at equilibrium. Therefore, the sum of the free energies of the products and the reactants is same at equilibrium
Hence, Gibbs energy change is zero for a reversible reaction at equilibrium is option (a). zero.
Note: We should know that the total free energy of the system (reactants + products) always becomes more negative as the reaction proceeds. Eventually it reaches a minimum value at a system composition that defines the equilibrium composition of the system, after which time no further net change will occur.
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