Methanol has the formula of CH3OH and can be produced by the reaction of carbon monoxide with hydrogen gas. CO + 2 H2 === CH3OH + heat Using the above chemical reaction, determine how each of the following actions would affect the equilibrium of the system. a. heating the mixture b. adding an excess of carbon monoxide c. removing the methanol as it is formed d. adding a substance that reacts with carbon monoxide

See explanation below

Explanation:

Let's answer this by parts.

a) As the reaction is naturally exothermic (Exerts heat), increasing temperature means that the reaction will move forward the direction where its absorbing heat

, so the equilibrium will be favored to the reactant side.

b) In this case, the addition of CO, which is a reactant will favor the reaction to the product side. This is because putting more quantity in the reactants will cause that the equilibrium constant K, increase it's innitial value, and this favors the product side.

c) In this case, it happens something similar to case b. This will cause that the reaction will have to work to cover the loss of methanol, therefore, equilibrium will move to the product side too.

d) In this case, we are adding another substance that reacts with CO, so it means that the other reactant is being removed, therefore, the reaction will work to cover this loss but also to cover the adding of the new substance, causing that the K decrease it's value, so equilibrium will be favored to the reactants side.

Hope this helps

the right answer is c i just did the question.