Chemistry, 20.03.2020 09:47 jessicamcummins

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:

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Answer from: Jaymp2706jj

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

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 from: Quest

Adozen is 12, and your needing to calculate the weight of one medium egg, so i would do 21/12=1.75 per egg, and check your answer by doing 1.75 * 12= 21 : ) : ): ): ): ) i hoped i

Answer from: Quest

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

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This reaction has an equilibrium constant of Kp = 2.26 * 104 at 298 K: CO(g) + 2 H2(g) ∆ CH3OH(g) Ca...