# In the following equation: if you add CH_4 to the mixture which direction does it shift? If you remove H_2 what direction does it shift? Which way does it shift if you add a catalyst?

## Equilibrium equation: $C {H}_{4} + {H}_{2} O \to C {H}_{3} O H + {H}_{2} + h e a t$. All reactants and products are gases.

Sep 17, 2016

Adding $C {H}_{4}$ or removing ${H}_{2}$ will both shift the reaction to the right (more product formation). Catalysts only increase the rate of reaction, not the direction.

#### Explanation:

Equilibrium reaction equations can be looked at like balances – or seesaws – where the amounts of reactants (left side) want to ‘balance’ with the amounts of products (right side) according to the relevant equilibrium reaction constants for the compounds.

So, when you add $C {H}_{4}$ on the left, the “seesaw” tips to the left. Thus, to ‘balance’ it again, something needs to move to the right side. Similarly, removing ${H}_{2}$ ‘lightens’ the right side. Again, to return to a balance, something needs to move to the right side.

Generally, adding more reactant or removing a product shifts the equilibrium to the right – more product needs to be generated. Removing a reactant or allowing the product to accumulate (or adding more of it to the reaction) shifts the equilibrium to the left – more reactants will remain in the system.