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,
[tex]R(g)+2T(g)\rightleftharpoons 2X(g)+Z(g)[/tex]
The expression of [tex]K_c[/tex] will be,
[tex]K_c=\frac{[Z][X]^2}{[R][T]^2}[/tex]
where,
[tex]K_c[/tex] = 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:
[tex]16=\frac{(2.0)\times (2.0)^2}{(2.0)\times [T]^2}[/tex]
[tex][T]=0.5M[/tex]
Therefore, the equilibrium concentration of T(g) is 0.5 M
