Question

For a chemical reaction, \[{K_{eq}}\]is 100 at 300K, and the value of \[\Delta G^\circ \] is –xR Joule at 1 atm pressure. Find the value of x. (Use ln 10 = 2.3)

Answer 1

Hint: We know that equilibrium constant and Gibbs free energy change can be utilised to foreknow the ratio of products to reactants at equilibrium for a given reaction. Gibbs free energy change is linked to the equilibrium constant by the equation\[\Delta G^\circ = - RTlnK\].

Formula Used:
\[\Delta G^\circ = - RTlnK\]
\[\Delta G^\circ \]= Gibbs free energy change
\[{K_{eq}}\]=equilibrium constant
T=temperature
R=universal gas constant

Complete Step by Step Answer:
Here in this question, we are given the equilibrium constant and Gibbs free energy change at a particular temperature.
The value of Gibbs free energy contains a variable. We have to find out the value of that variable i.e., x.
Along with it, we are given a temperature also.
We know the relationship between \[\Delta G^\circ \]and \[{K_{eq}}\].
By using this relationship we can get the answer.
\[\Delta G^\circ = - RTlnK\]
\[ \Rightarrow - xR = - R\left( {300} \right)ln\left( {100} \right)\]
Cancelling R both sides we get,
\[ \Rightarrow - x = \left( {300K} \right)ln\left( {100} \right)\]
We can write log(100) as log(\[{10^2}\]) which is equal to 2log(10).
Putting this value in the above equation we get,
\[ \Rightarrow x = \left( {300} \right)\left( {2log10} \right)\]
\[ \Rightarrow x = \left( {300} \right)2\left( {2.3} \right)\]
\[ \Rightarrow x = 1380\]
So, \[x = 1380\].

So the value of x is 1380.

Additional Information: If a system is not at equilibrium,\[\Delta G\], and reaction quotient, Q can be utilised to inform the direction of the reaction that must move to achieve equilibrium.
\[\Delta G\]is associated with Q by the equation \[\Delta G = RTln\frac{Q}{K}\].
If \[\Delta G < 0\], then \[K > Q\], and the reaction must move to the right to achieve equilibrium.
If \[\Delta G > 0\], then\[K < Q\], and the reaction must move to the left to achieve equilibrium.
If \[\Delta G = 0\], then\[K = Q\], and the reaction has achieved equilibrium.

Note: If \[\Delta G^\circ < 0\], then \[K > 1\], and products will be preferred over reactants at equilibrium. If\[\Delta G^\circ > 0\], then\[K < 1\], and reactants are preferred over products at equilibrium. If\[\Delta G^\circ = 0\], then \[K = 1\], and a considerable no.of products will be approximately equal to no.of considerable reactants at equilibrium. This is an infrequent event for chemical reactions.