Q.Describe the effect of: a) addition of H2 b) addition of CH3OH c) removal of CO d) removal of CH3OH on the equilibrium of the reaction: 2H2(g) + CO

The key idea is Le Chatelier’s Principle: a system at equilibrium shifts to counteract any change in concentration.

a) Addition of $H_2$ increases the concentration of a reactant. The equilibrium shifts forward (towards products) to consume the added $H_2$, producing more $CH_3OH$.

b) Addition of $CH_3OH$ increases the concentration of the product. The equilibrium shifts backward (towards reactants) to consume the added $CH_3OH$, producing more $H_2$ and $CO$.

c) Removal of $CO$ decreases the concentration of a reactant. The equilibrium shifts backward (towards reactants) to replace the lost $CO$, consuming $CH_3OH$. …

The core idea here is beautifully simple: a system at equilibrium “fights back” against any disturbance. If you add something, it tries to use it up. If you remove something, it tries to make more of it. This is Le Chatelier’s Principle, and it’s all you need for this question.

Let’s look at the reaction carefully:

Notice the stoichiometry: two moles of hydrogen gas and one mole of carbon monoxide gas combine to form one mole of methanol gas. This matters because the “shift” in equilibrium isn’t just about adding or removing stuff — it’s about which side has more molecules. But here, we’re only changing concentrations, so we focus on that.

1. Addition of $H_2$ (a)

   You’re adding a reactant. The system has too much hydrogen now, so it tries to consume it. The only way to do that is to push the reaction forward — toward the product side. So the equilibrium shifts to the right, producing more $CH_3OH$.

   Tip

   Adding any reactant (or product) always shifts equilibrium away from that side. Think of it as the system “using up” the extra.

2. Addition of $CH_3OH$ (b)

   Now you’re adding a product. The system has too much methanol, so it tries to get rid of it. It does this by shifting the reaction backward — toward the reactants. So the equilibrium shifts to the left, producing more $H_2$ and $CO$.

   Watch out

   A common mistake is to think adding a product always shifts right. It doesn’t — it shifts away from the side you added to.

3. Removal of $CO$ (c) …