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?
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.
For number four the answer would be reactant? Correct?
1) To determine which side of the reaction will be favored by a decrease in pressure, you need to consider the number of moles of gas on each side of the reaction. In this reaction, there are 4 moles of gas on the reactant side and 4 moles of gas on the product side.
When the pressure is decreased, the equilibrium will shift in the direction that reduces the number of moles of gas. In this case, both the reactant and product sides have the same number of moles of gas, so a decrease in pressure will not favor either side. Therefore, the answer is "neither."
2) The removal of O2 will lead to a decrease in the concentration of O2 in the reaction mixture. According to Le Chatelier's principle, when a reactant is removed, the equilibrium will shift in the direction that replenishes the reactant.
In this case, the removal of O2 will shift the equilibrium to the left side to produce more O2. Therefore, the reactant side will be favored. So, the answer is "reactant."
3) Similar to the previous question, if you decrease the amount of C(s), the equilibrium will shift to replenish the reactant. In this case, that means the equilibrium will shift to the left side to produce more C(s). Therefore, the reactant side will be favored. So, the answer is "reactant."
4) To determine which side of the reaction will be favored by a decrease in pressure, you need to consider the number of moles of gas on each side of the reaction. In this reaction, there are 2 moles of gas on the reactant side and 1 mole of gas on the product side.
When the pressure is decreased, the equilibrium will shift in the direction that reduces the number of moles of gas. In this case, reducing the pressure will favor the side with fewer moles of gas, which is the product side (N2O4). Therefore, the answer is "product."