Question

At 298K, the equilibrium constant for the following reaction is 1.00 107 H2S(aq) + H2O- H30"(aq) + HS-(aq) The equilibrium constant for a second reaction is 1.00 10-19. Use this information to determine the equilibrium constant for the reaction: (aq)+2H2O 2H3O K- Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

The equilibrium constant is the ratio of concentration of products to the concentration of reactants.

For forward reaction if the equilibrium constant is Kq, then for reverse reaction the equilibrium constant is 1/Kq.

If we are adding to reactions of equilibrium constantsK1 and K2 respectively, then the constant for the resultant reaction will be K1.K2.

If we are substracting 2 reactions then the equilibrium constant for the resultant reaction will be

KI/K2.

$H_2S+H_2O\rightleftharpoons H_3O^++HS^-\rightarrow 1;K_1$

$HS^-+H_2O\rightleftharpoons H_3O^++S^{2-}\rightarrow 2;K_2$

If we add the equations 1 and 2 we get

$H_2S+2H_2O\rightleftharpoons 2H_3O^++S^{2-}\rightarrow 3$

So the equilibrium constant for reaction 3 will be K1.K2.

It will be equal to $1*10^{-26}$ .