Question

The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2 mole of x and 3 mole of z in a one litre vessel?​

Answer 1

To find the equilibrium concentration of y, we use the equilibrium constant expression K = [z]^2 / ([x]^2[y]), plug in the known values and solve for [y], which results in the concentration of y at equilibrium being 1 M.

Step-by-step explanation:

The student is asking about the concentration of gas y at equilibrium when the equilibrium constant (K) for the reaction 2x(g) + y(g) = 2z(g) is given as 2.25, and the initial concentrations of x and z are 2 M and 3 M, respectively, in a 1-liter vessel. To find the concentration of y at equilibrium, we use the expression for the equilibrium constant K which is:

K = [z]2 / ([x]2[y])

Plugging in the known values and solving for [y] we get:

2.25 = (3)2 / (22[y])

[y] = (32) / (2.25 × 22)

[y] = 9 / 9

[y] = 1 M

Thus, the concentration of y at equilibrium is 1 M in the 1-liter vessel.

Answer 2

`[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^(-3)` at equilibrium.

Explanation

Concentration for each of the species:

• 
  `[\text{X}] = (n)/(V) = 2\; \text{mol}\cdot \text{dm}^(-3)`;
• 
  `[\text{Y}] = (n)/(V) = 0\; \text{mol}\cdot \text{dm}^(-3)`;
• 
  `[\text{Z}] = (n)/(V) = 3\; \text{mol}\cdot \text{dm}^(-3)`.

There was no Y to start with; its concentration could only have increased. Let the change in
`[\text{Y}]` be
`+x \; \text{mol}\cdot \text{dm}^(-3)`.

Make a
`\textbf{RICE}` table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the change in
`[\text{X}]` will be
`+2\;x \; \text{mol}\cdot \text{dm}^(-3)` and the change in
`[\text{Z}]` will be
`-2\;x \; \text{mol}\cdot \text{dm}^(-3)`.

.

Add the value in the C row to the I row:

.

What's the equation of
`K_c` for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

`K_c = \frac{[\text{Z}]^(2)}{[\text{X}]^(2) \cdot[\text{Y}]}`.

`K_c = 2.25`. As a result,

`\frac{[\text{Z}]^(2)}{[\text{X}]^(2) \cdot[\text{Y}]} = ((3-2x)^(2))/((2+2x)^(2) \cdot x) = K_c = 2.25`.

`(3-2\;x)^(2)= 2.25 \cdot(2+2\;x)^(2) \cdot x\\4\;x^(2) - 12 \;x + 9 = 2.25 \;(4\;x^(3) + 8 \;x^(2) + 4 \;x)\\4\;x^(2) - 12\;x + 9 = 9 \;x^(3) + 18\;x^(2) + 9\;x\\9\;x^(3) + 14\;x^(2) + 21\;x - 9 = 0`.

The degree of this polynomial is three. Plot the equation
`y = 9\;x^(3) + 14\;x^(2) + 21\;x - 9` on a graph and look for any zeros. There's only one zero at
`x \approx 0.337`. All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y:
`0.337\;\text{mol}\cdot\text{dm}^(-3)`.