H2(g) + I2(g) 2 HI(g)
The forward reaction above is exothermic. At equilibrium, what happens if I2 is removed from the reaction mixture at constant temperature and volume? select all that apply
1.) Energy is absorbed
2.) Energy is released
3.) [H2] increases
4.) [H2] decreases
5.) [H2] remains constant
6.) [HI] increases
7.) [HI] decreases
8.) [HI] remains constant
Hey there! Let me give you some clownish answers to your question:
Alright, so the forward reaction is exothermic, which means it releases energy.
Now, if we remove I2 from the reaction mixture at constant temperature and volume...
Well, let's think about it together:
Since I2 is part of the reactants, by removing it, we are disturbing the equilibrium. To counteract the disturbance, the reaction would shift in the direction that opposes the change.
So, here are the effects we can expect:
1.) Energy is absorbed: Nope! The forward reaction is exothermic, so removing a reactant wouldn't cause the energy to be absorbed.
2.) Energy is released: Correct! Removing I2 would shift the reaction to produce more HI, which is the exothermic reaction, releasing more energy.
3.) [H2] increases: Yup! The reaction would shift to increase the concentration of H2 since it's part of the reactants.
4.) [H2] decreases: Nah, wrong clown car! H2 concentration would actually increase in this case.
5.) [H2] remains constant: Nope! As mentioned earlier, [H2] would increase.
6.) [HI] increases: Exactly right! Removing I2 would shift the reaction to produce more HI.
7.) [HI] decreases: Nope, incorrect! Removing I2 would encourage the reaction to produce more HI.
8.) [HI] remains constant: Wrong answer, my friend! Removing I2 would definitely cause [HI] to increase.
So, the correct answers are 2, 3, and 6.
I hope my clownish answer brought a smile to your face!
When I2 is removed from the reaction mixture at constant temperature and volume, the following changes occur:
2.) Energy is released
4.) [H2] decreases
6.) [HI] increases
8.) [HI] remains constant
To analyze the effect of removing I2 from the reaction mixture, we need to consider Le Chatelier's principle.
1. Energy is absorbed: Removing I2 will disturb the equilibrium, but it won't directly affect the energy of the system.
2. Energy is released: Since the forward reaction is exothermic, removing I2 would reduce the amount of HI being formed, which means less heat would be released.
3. [H2] increases: Removing I2 would shift the equilibrium towards the reactants to restore equilibrium. As a result, the concentration of H2 would increase.
4. [H2] decreases: This is incorrect. The removal of I2 would result in a shift towards the reactants, causing an increase in [H2], not a decrease.
5. [H2] remains constant: This is incorrect. Removing I2 would cause a shift towards the reactants, increasing the concentration of H2.
6. [HI] increases: Removing I2 would cause the equilibrium to shift towards the reactants. As a result, more HI would be produced to restore equilibrium.
7. [HI] decreases: This is incorrect. The removal of I2 would result in a shift towards the reactants, causing an increase in [HI], not a decrease.
8. [HI] remains constant: This is incorrect. Removing I2 would cause a shift towards the reactants, increasing the concentration of HI.
Therefore, the correct answers are:
- Energy is released
- [H2] increases
- [HI] increases