In a systemAB(s)<-------->A (g)+ B(g)doubling the quantity of AB(s) what would be change in other side of reactiom

in the given reaction AB(s)<-------->A (g)+ B(g) [AB]=1, because of solid phase [A]= concentration of A(gas) [B]= concentration of B(gas) Equilibrium constact

AB(s)<-------->A (g)+ B(g) The concentration of AB will be unity because of solid. So [AB]=1 This reaction is in equilibrium so on doubling of AB(s), it will not go in the forward direction because of solid phase. The double amount will be as solid. So there will not be any change in the gases concentration. The reaction will not move in forward and backward direction and there will not be any change in the other side of the reaction This question is based on the concept of reactivity of a compound in heterogeneous equilibrium reaction.

there will b no change in  the forward reaction as active mass of solids is constant and concentration of non gaseous substances is not written k_c  or k_p expression.