Question
N2O4(g) + energy ⇄ 2NO2(g)
The system is at equilibrium, but then it is subjected to an increase in pressure. Which change will happen?
(1 point)
Responses
The rate of the reverse reaction will increase, which will increase the number of particles in the system.
The rate of the reverse reaction will increase, which will increase the number of particles in the system.
The rate of the forward reaction will increase, which will decrease the number of particles in the system.
The rate of the forward reaction will increase, which will decrease the number of particles in the system.
The rate of the forward reaction will increase, which will increase the number of particles in the system.
The rate of the forward reaction will increase, which will increase the number of particles in the system.
The rate of the reverse reaction will increase, which will decrease the number of particles in the system.
The system is at equilibrium, but then it is subjected to an increase in pressure. Which change will happen?
(1 point)
Responses
The rate of the reverse reaction will increase, which will increase the number of particles in the system.
The rate of the reverse reaction will increase, which will increase the number of particles in the system.
The rate of the forward reaction will increase, which will decrease the number of particles in the system.
The rate of the forward reaction will increase, which will decrease the number of particles in the system.
The rate of the forward reaction will increase, which will increase the number of particles in the system.
The rate of the forward reaction will increase, which will increase the number of particles in the system.
The rate of the reverse reaction will increase, which will decrease the number of particles in the system.
Answers
GPT-4o mini
To determine the effect of an increase in pressure on the equilibrium of the reaction
\[ \text{N}_2\text{O}_4(g) + \text{energy} \rightleftharpoons 2\text{NO}_2(g), \]
we need to apply Le Chatelier's principle, which states that if a system at equilibrium is subjected to a change in pressure, temperature, or concentration, the system will adjust to counteract that change and restore a new equilibrium.
In this case, when pressure is increased, the system will shift in the direction that produces fewer gas molecules.
- On the left side (reactants), we have 1 mole of N₂O₄.
- On the right side (products), we have 2 moles of NO₂.
Since there is 1 mole of gas on the left and 2 moles on the right, increasing the pressure will shift the equilibrium to the left (toward the reactants) to reduce the number of gas molecules.
Therefore, the rate of the reverse reaction (which converts NO₂ back into N₂O₄) will increase, which will result in a decrease in the number of gas particles in the system.
Based on this analysis, the correct response is:
**The rate of the reverse reaction will increase, which will decrease the number of particles in the system.**
\[ \text{N}_2\text{O}_4(g) + \text{energy} \rightleftharpoons 2\text{NO}_2(g), \]
we need to apply Le Chatelier's principle, which states that if a system at equilibrium is subjected to a change in pressure, temperature, or concentration, the system will adjust to counteract that change and restore a new equilibrium.
In this case, when pressure is increased, the system will shift in the direction that produces fewer gas molecules.
- On the left side (reactants), we have 1 mole of N₂O₄.
- On the right side (products), we have 2 moles of NO₂.
Since there is 1 mole of gas on the left and 2 moles on the right, increasing the pressure will shift the equilibrium to the left (toward the reactants) to reduce the number of gas molecules.
Therefore, the rate of the reverse reaction (which converts NO₂ back into N₂O₄) will increase, which will result in a decrease in the number of gas particles in the system.
Based on this analysis, the correct response is:
**The rate of the reverse reaction will increase, which will decrease the number of particles in the system.**