Question

This reaction has an equilibrium constant of Kp = 2.26 * 104 at 298 K: CO(g) + 2 H2(g) ∆ CH3OH(g) Calculate Kp for each reaction and predict whether reactants or products will be favored at equilibrium:

Answer 1

This reaction has an equilibrium constant of $K_p=2.26\times 10^4$ at 298K.

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

Calculate Kp for each reaction and predict whether reactant or products will be favoured at equilibrium.

A) $CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)$

B) $\frac{1}{2}CO(g)+H_2(g)\rightleftharpoons \frac{1}{2}CH_3OH(g)$

C) $2CH_3OH(g)\rightleftharpoons 2CO(g)+4H_2(g)$

Answer:

A) $CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)$ : $0.442\times 10^-4}$ : reactants are favoured

B) $\frac{1}{2}CO(g)+H_2(g)\rightleftharpoons \frac{1}{2}CH_3OH(g)$: $1.50\times 10^2}$ : products are favoured

C) $2CH_3OH(g)\rightleftharpoons 2CO(g)+4H_2(g)$ : $4.42\times 10^{-9}$: reactants are favoured

Explanation:

1. We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

$CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)$

$K_{p}'=(\frac{1}{K_p})$

$K_{p}'=(\frac{1}{2.26\times 10^4})=0.442\times 10^-4}$

As $K_p'$ is less than 1, reactants will be favored at equilibrium.

2. We need to calculate the equilibrium constant for the half equation of above chemical equation, which is:

$\frac{1}{2}CO(g)+H_2(g)\rightleftharpoons \frac{1}{2}CH_3OH(g)$

$K_{p}'=(\sqrt{K_p})$

$K_{p}'=(\sqrt{2.26\times 10^4})=1.50\times 10^2}$

As $K_p'$ is greater than 1, products will be favored at equilibrium.

3.  We need to calculate the equilibrium constant for the reverse and twice the equation of above chemical equation, which is

$2CH_3OH(g)\rightleftharpoons 2CO(g)+4H_2(g)$

$K_{p}'=(\frac{1}{K_p})^2$

$K_{p}'=(\frac{1}{2.26\times 10^4})^2=4.42\times 10^{-9}$

As $K_p'$ is less than 1, reactants will be favored at equilibrium.

Answer 2

The equilibrium constant for the given reaction has been $2.26\;\times\;10^4$. The value of equilibrium constant has been delivering to whether the reactant has been favored or product.

• The equilibrium constant for, $\rm CH_3OH\leftrightharpoons CO\;+\;H_2$.

The reaction has been the reverse of the given reaction. The equilibrium constant has been given by:

$Kp'=\dfrac{1}{Kp}\\Kp'= \dfrac{1}{2.26\;\times\10^4}\\Kp'=0.442\;\times\;10^-^4$

The value of equilibrium constant has been $\rm 0.442\;\times\;10^-^4$. The value has been less than 1, thus favors reactant formation.

• The equilibrium constant for, $\rm \dfrac{1}{2}\;CO\;+\;H_2 \;\leftrightharpoons \dfrac{1}{2}\;CH_3OH$

The reaction has been half of the given reaction. The equilibrium constant has been given by:

$Kp'=\sqrt{Kp} \\Kp'=\sqrt{2.26\;\times\;10^4} \\Kp'=1.50\;\times\;10^2$

The value of equilibrium constant has been $\rm 1.5\;\times\;10^2$. The value has been more than 1, thus favors product formation.

• The equilibrium constant for, $\rm 2\;CH_3OH\leftrightharpoons 2\;CO\;+\;4\;H_2$

The reaction has been reverse and double of the given reaction. The equilibrium constant has been given by:

$Kp'=\dfrac{1}{Kp}^2\\Kp'= (\dfrac{1}{2.26\;\times\;10^4})^2\\Kp'=4.42\;\times\;10^-^9$

The value of equilibrium constant has been $\rm 4.42\;\times\;10^-^9$. The value has been less than 1, thus favors reactant formation.

For more information about equilibrium constant, refer to the link:

https://brainly.com/question/17960050