I don't understand the effect of temperature on a reaction. For exmaple:

Question

CO + 3H2 <-> CH4 + H2O deltaH= -230 KJ

If there were an increase in temperature then the CH4 and H2O would have a harding time becoming the CO and H2 so therefore the reaction would move to the right. Is that right?

Answer 1

Here is a tip. Always replace the -delta H or + delta H like this.
CO + 3H2 <==>CH4 + H2O + heat (if delta H were + it means the reaction is endothermic instead of exothermic and I would have written heat on the left side with the reactants. Now that the heat is placed properly, just answer as you would for any change in concentration of the molecules. So adding heat to the reacting makes the reaction shift so as to use up the heat and that means the reaction shifts to the left. (You see that CH4 + H2O + heat makes the reactants and heat is used to do that so the reaction moves to the left.)

Answer 2

Ohh wow, what an awesome tip! Thank you so much. This was so confusing until now :D

So for something like (losing heat)
2A + B <-> 2C delta H postive
it would be
2A + B <-> 2C -heat
the reaction would move right?

Answer 3

Yes, but since negative signs can be confusing, I would write it as
2A + B + heat <==> 2C, then adding heat makes the reaction shift to the right. That way the rules I follow stay the same; i.e., it moves AWAY from what is being added. So adding A or B or heat would make it move to the right; adding C or cooling it would make it move to the left

Answer 4

Thank You :)