Question

A 1.0 mol sample of X(g) and a 1.0 mol sample of Q(g) are introduced into an evacuated, rigid 10.0 L container and allowed to reach equilibrium at 50ºC according to the equation above. At equilibrium, which of the following is true about the concentrations of the gases?
X(g) + 2 Q(g) --> R(g) + Z(g)

Kc = 1.3 x 105 at 50ºC

[R] = ½ [Q]
[X] = [Q] = [R] = [Z]
[Q] = ½ [X]
[R] = [Z] > Q

Answer 1

Answer:

[R] = [Z] > Q

Explanation:

Given the equilibrium:

                        X(g) + 2 Q(g) ⇄  R(g) + Z(g)

The equilibrium constant will be given by:

Kc = 1.3 x 105 at 50ºC =  [R] x [Z] / ( [X] x [Q]² )

This value is telling us that at equilibrium the products quantities will be much higher than the reactants.

Looking at the answer choices we see that (a) is false since this is an equilibrium not a areaction that goes to completion;(b) is is false because again this is an equilibrium and it would also imply that Kc is one; (c) is false because in the reaction the ratio is 1X : 2Q.

The last choice is the true statement since the equilibrium constant favors the products. This statement would haven true if Kc had been much less than one.

Answer 2

Answer: At equilibrium, the true statement is [R] = [Z] > [Q]

Explanation:

We are given:

Initial moles of X = 1.0 moles

Initial moles of Q = 1.0 moles

Volume of container = 10.0 L

Molarity is calculated by using the formula:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of container}}$

$\text{Initial molarity of X}=\frac{1.0}{10.0}=0.1M$

$\text{Initial molarity of Q}=\frac{1.0}{10.0}=0.1M$

For the given chemical equation:

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

Initial:          0.1         0.1

At eqllm:    0.1-x      0.1-2x     x       x

The expression of $K_c$ for above equation follows:

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

We are given:

$K_c=1.3\times 10^5$

Putting values in above expression, we get:

$1.3\times 10^5=\frac{x\times x}{(0.1-x)\times (0.1-2x)}\\\\x=0.049,0.1$

Neglecting the value of x = 0.1 because the concentration of Q will become negative, which is not possible

So, equilibrium concentration of X = (0.1 - x) = (0.1 - 0.049) = 0.051 M

Equilibrium concentration of Q = (0.1 - 2x) = [0.1 - 2(0.049)] = 0.002 M

Equilibrium concentration of R = x = 0.049 M

Equilibrium concentration of Z = x = 0.049 M

Hence, at equilibrium, the true statement is [R] = [Z] > [Q]