Question

The reaction mixture represented above is at equilibrium at 298 K. The value of the equilibrium constant for the reaction is 16 at 298 K. If R(g), X(g) and Z(g) each have an equilibrium concentration of 2.0 M, what is the equilibrium concentration of T(g)?

Answer 1

Answer : The equilibrium concentration of T(g) is 0.5 M

Solution :

Let us assume that the equilibrium reaction be:

The given equilibrium reaction is,

$R(g)+2T(g)\rightleftharpoons 2X(g)+Z(g)$

The expression of $K_c$ will be,

$K_c=\frac{[Z][X]^2}{[R][T]^2}$

where,

$K_c$ = equilibrium constant = 16

[Z] = concentration of Z at equilibrium = 2.0 M

[R] = concentration of R at equilibrium = 2.0 M

[X] = concentration of X at equilibrium = 2.0 M

[T] = concentration of T at equilibrium = ?

Now put all the given values in the above expression, we get:

$16=\frac{(2.0)\times (2.0)^2}{(2.0)\times [T]^2}$

$[T]=0.5M$

Therefore, the equilibrium concentration of T(g) is 0.5 M