For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbls energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to zero.

Which diagram represents the large value of equilibrium constant for the reversible reaction

For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbls energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to zero. The value of "log"_(10) K_(eq) is equal to [ k_(eq) is the equilibrium coustant]

For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbls energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to zero. For a reaction M_(2)O_((s))^(3//4) rarr 2 M_((S)) + (1)/(2) O_(2 (g)) Delta H = 30 KJ//mol and Delta S = 0.07 KJ//mol //K at 1 atm. The reaction would not be spontaneous at temperature

How does a catalyst influence the equilibrium constant of a reversible reaction?

For a reversible reaction if the concentration of the reactants are doubled, the equilibrium constant will be :

With increase in temperature generally the value of the equilibrium constant of endothermic reversible reaction

With increase in temperature generally the value of the equilibrium constant of endothermic reversible reaction

Write the equilibrium expression and equilibrium constant for the given reverse reaction.

In a reversible reaction, if the concentration of reactants are doubles, the equilibrium constant K will: