Here is what you need to know.
For gaseous reactions, an INCREASE in P shifts the reaction to the side with fewer moles (because it is trying to occupy a smaller volume as P gets larger). For #1, you have 5 moles on the left, 4 on the right; therefore, an increase in P will shift the equilibrium to the right (products favored). A decrease must do the opposite.
#2 is correct.
#3. C is a solid. Decreasing (or increasing ) C will not affect the reaction as long as there is at least 1 molecule of C in the reaction vessel. So the answer is neither.
#4. Now that you know how to answer #1 you should be able to do #4 easily.
I have 4 questions, I got the answer for three of them and just want you to check and see if they are right and i do not know the answer to the fourth one.
1)4HCl(g)+O2(g)<=>2H2O(g)+2Cl2(g)+114.4 kJ Which side of the reaction will be favoured by an decrease in pressure ? The answer can be product, reactant or neither and i think the answer is product?
2)4HCl(g)+O2(g)<=>2H2O(g)+2Cl2(g)+114.4 kJ Which side of the reaction will be favoured by the removal of O2? Again the answer can be product, reactant or neither and i think it is reactant?
3)C(s)+H2O(g)<=>CO(g)+H2(g)+114.4 kJ Which side of the reaction will be favoured by a decrease in the amount of C(s)? The answer can be product, reactant or neither and i think it is reactant?
4)2NO2(g)<=>N2O4(g)+58kJ Which side of the reaction will be favoured by an decrease in pressure ? The answer can be product, reactant or neither?
2 answers
For number four the answer would be reactant? Correct?