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

To find the equilibrium concentration of T(g) when R(g), X(g), and Z(g) each have an equilibrium concentration of 2.0 M, and the equilibrium constant is 16, we solve the equilibrium expression Ke = [T][Z]/[R][X] to obtain T(g) = 1.0 M.

Step-by-step explanation:

The student's question pertains to chemical equilibrium and involves the calculation of equilibrium concentrations using the equilibrium constant (Ke) and stoichiometry of the reaction. The problem involves an equilibrium system at 298 K with known concentrations of reactants R(g), X(g), Z(g), and the value of Ke being 16. Assuming the reaction is R(g) + X(g) <=> T(g) + Z(g), we can set up an ICE (Initial, Change, Equilibrium) table to calculate for T(g) concentration. Since the equilibrium concentrations of R, X, and Z are given as 2.0 M, we can express the change in concentration of T(g) as +x and for R(g) and X(g) as -x. Hence, if we assume that the concentration of T(g) initially was zero, the equilibrium concentration for T(g) will be simply x.

Using the equilibrium expression for the reaction, Ke = [T][Z]/[R][X], where the concentration of T(g) is x, and the concentrations of R(g), X(g), and Z(g) are 2.0 M, we can solve for x:

Ke = (x)(2.0) / (2.0-x)(2.0-x) = 16

Simplifying and solving for x, we get x = 1.0 M assuming that subtraction of x does not significantly alter the concentrations of R and X due to the relatively large equilibrium constant. Thus, the equilibrium concentration of T(g) is 1.0 M.

Answer 2

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=([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=((2.0)* (2.0)^2)/((2.0)* [T]^2)`

`[T]=0.5M`

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