I really appreciate the help you gave me earlier. I was wondering if you could please look over these questions for me to? The answers will always be either reactant, product or neither.
1)P4(s)+6 Cl2(g)<=>4 PCl3(l)
Which side of the reaction will be favoured by an increase in pressure ?
I chose product?
2) P4(s)+6 Cl2(g)<=>4 PCl3(l)
Which side of the reaction will be favoured by an increase in the amount of PCl3(l) ? I chose reactant?
3) P4(s)+6 Cl2(g)<=>4 PCl3(l)
Which side of the reaction will be favoured by a decrease in the amount of Cl2(g) ? I think it is the product?
4) 2 NO2(g)<=>N2O4(g)+58 kJ
Which side of the reaction will be favoured by the removal of NO2 ?
I think reactant?
5)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 CO(g) ? I think it is product?
6) 2 NO2(g)<=>N2O4(g)+58 kJ
Which side of the reaction will be favoured by the addition of an inhibitor ? I think it is neither?
Thanks again for the help. I appreciate it!
2 is neither. Adding PCl3 does not change the concn. Solids, if that is their normal state as in the carbon question before, and liquids (if that is their normal state) are not included in Keq.
3 is the opposite to your answer for the same reasons that your 4 and 5 are correct.
Le Chatelier's Principle, which I assume you are studying, says this is VERY common terms. When a system in equilibrium is subjected to a stress, it will shift (move one direction or the other) to undo what we did to it.
So if we add Cl2 to #3 it will shift so as to undo what we did. What is that? We added so it will try to take it away. And the way it can take away Cl2 is by shifting to the right which means the products are favored. Decreasing a substance is just the opposite; i.e., it will shift to make more of what we've taken away. In the case of #3, removing Cl2 will make it shift to the left and the reactants are favored.
Sure, I'd be happy to look over your questions and provide feedback on your answers.
1) P4(s) + 6 Cl2(g) <=> 4 PCl3(l)
Question: Which side of the reaction will be favored by an increase in pressure?
Your answer: Product
Explanation: In general, increasing pressure favors the side of the reaction with fewer gas molecules. In this case, the left side (reactants) has 7 gas molecules (1 for P4 and 6 for Cl2), while the right side (product) has 4 gas molecules (all PCl3). Therefore, an increase in pressure would favor the side with fewer gas molecules, which is the product side. So your answer is correct.
2) P4(s) + 6 Cl2(g) <=> 4 PCl3(l)
Question: Which side of the reaction will be favored by an increase in the amount of PCl3(l)?
Your answer: Reactant
Explanation: Increasing the amount of a product will shift the equilibrium towards the reactant side so that more reactant can be converted into product until a new equilibrium is reached. Therefore, your answer is correct.
3) P4(s) + 6 Cl2(g) <=> 4 PCl3(l)
Question: Which side of the reaction will be favored by a decrease in the amount of Cl2(g)?
Your answer: Product
Explanation: Removing Cl2(g) will shift the equilibrium towards the Cl2(g)-containing side in order to restore the balance. So, a decrease in the amount of Cl2(g) will favor the reactant side. Therefore, your answer is incorrect. It should be the reactant side.
4) 2 NO2(g) <=> N2O4(g) + 58 kJ
Question: Which side of the reaction will be favored by the removal of NO2?
Your answer: Reactant
Explanation: Removing NO2 will shift the equilibrium towards the side with more NO2 molecules to restore the balance. Therefore, your answer is correct.
5) C(s) + H2O(g) <=> CO(g) + H2(g) + 114.4 kJ
Question: Which side of the reaction will be favored by a decrease in the amount of CO(g)?
Your answer: Product
Explanation: Decreasing the amount of a product will shift the equilibrium towards the reactant side to produce more product until a new equilibrium is reached. Therefore, your answer is incorrect. It should be the reactant side.
6) 2 NO2(g) <=> N2O4(g) + 58 kJ
Question: Which side of the reaction will be favored by the addition of an inhibitor?
Your answer: Neither
Explanation: An inhibitor does not selectively favor any side of the reaction. Instead, it works by slowing down the reaction rate by interfering with the reaction mechanism. Therefore, your answer is correct.
Overall, you did a great job answering most of the questions correctly. Keep up the good work! If you have any more questions, feel free to ask.