Question

3a(g)+b(g)⇌2c(g) Initial moles of A is taken as 10, Kc is given as 8000. What is the number of moles of A,B and C at equilibrium

Answer 1

Answer:

The equilibrium reaction is 3A(g)+B(g)⇌2C(g).

The equilibrium reaction is 3A(g)+B(g)⇌2C(g).Let V be the volume of the flask.

The equilibrium reaction is 3A(g)+B(g)⇌2C(g).Let V be the volume of the flask.The equilibrium number of moles of A,B and C are 2,2 and 2 respectively.

The equilibrium reaction is 3A(g)+B(g)⇌2C(g).Let V be the volume of the flask.The equilibrium number of moles of A,B and C are 2,2 and 2 respectively.The equilibrium concentrations of A,B and C are

respectively.

respectively.The expression for the equilibrium constant is K

respectively.The expression for the equilibrium constant is K c

respectively.The expression for the equilibrium constant is K c

respectively.The expression for the equilibrium constant is K c

respectively.The expression for the equilibrium constant is K c .

respectively.The expression for the equilibrium constant is K c .Substituting values in the above expression, we get

respectively.The expression for the equilibrium constant is K c .Substituting values in the above expression, we get9.0=

.

.Thus, V=6L.