The reaction which is in dynamic equilibrium, ensured us, that the reaction is reversible . But if that the reaction is in equilibrium. The reaction quotient predict either the reversible reaction is in equilibrium or tries to achieve equilibrium. In those reactions which have not achieved equilibrium, we obtain reaction quotient `Q_(c)` in place of equilibrium constant `(K_(c))` by substituting the concentration of reactant and product at the time, at whih we have to calculate the value of `Q_(c)` . To determine the direction at which the net reaction will proceed to achieve equilibrium, we compare values of `Q_(c)` and `K_(c)`. The three possible cases are shown as comparison of `K_(c)` and `Q_(c)` in the following figures.
Change in Gibbs free energy, i.e., `Delta G` is the driving force of any reaction.
For spontaneous reaction , `Delta G =-ve`
For non-spontaneous reaction , `Delta G=+ve`
For reaction at equilibrium , `Delta G =0`
Thermodynamically, we know that
`Delta G= Delta G^(@)+ RT ln Q`, where `Q` is reaction quotient and `Delta G^(@)=` change in Gibbs energy at standard condition.
For equilibrium `A(g) hArr B(g) (K_(eq) =1.732)` If the pressure of the system [varied by introducing a stream of `A (g)` and B (g) is represented by the curve at constant temperature T.
Suppose `N_(2)O_(4)(g)` is enclosed in a cylinder fitted with a movable piston which attains the following equilibrium
`N_(2)O_4)(g) hArr 2NO_(2)(g)`
Given that for the 10 atmosphere pressure of the equilibrium mixture, the content of `NO_(2)` is `8xx1^(5) ` ppm. if the piston of cylinder is moved upward in such a manner so that the volume of the gaseous mixture becomes double, then what will be new ppm of `NO_(2)(g)` in the cylinder ? ( Assuming that the temperature of the cylinder remains constant )
A. `8.2 xx 10^(5) `ppm
B. `10^(5)` ppm
C. `8.72 xx 10^(5)` ppm
D. `7.4 xx 10^(5)` ppm
