Question
In the reaction:
N2O4(g) + energy >< 2NO2(g). The system is at equilibrium but then gets an increase in pressure. Why will the increase in pressure cause the rate of the reverse reaction to increase and caused less particles in the system
N2O4(g) + energy >< 2NO2(g). The system is at equilibrium but then gets an increase in pressure. Why will the increase in pressure cause the rate of the reverse reaction to increase and caused less particles in the system
Answers
Answered by
GPT-4o mini
In the reaction you've provided:
\[ \text{N}_2\text{O}_4(g) + \text{energy} \rightleftharpoons 2\text{NO}_2(g) \]
we have one mole of dinitrogen tetroxide (N₂O₄) on the left side and two moles of nitrogen dioxide (NO₂) on the right side. When we talk about increasing pressure in a gaseous equilibrium, it's essential to apply Le Chatelier's Principle, which states that if a system at equilibrium experiences a change in pressure, temperature, or concentration, the system will adjust to counteract that change.
When pressure is increased in a gaseous system, the equilibrium will shift in the direction that produces fewer moles of gas. Since the left side (reactants) has 1 mole of N₂O₄ and the right side (products) has 2 moles of NO₂, increasing the pressure will favor the formation of N₂O₄ because it results in fewer total moles of gas in the system.
To break it down:
1. **Effect of Increased Pressure**: When the pressure is increased, the system will favor the side with fewer gas moles to reduce that pressure.
2. **Shifting to the Reverse Reaction**: In this case, the reaction will shift to the left (the reverse reaction) to produce more N₂O₄, which decreases the total number of gas molecules (from 2 moles of NO₂ to 1 mole of N₂O₄).
3. **Rate of Reverse Reaction**: The increase in pressure effectively increases the rate at which the reverse reaction occurs because the equilibrium position is shifting to the left, which means the system is producing more N₂O₄ from NO₂.
In summary, the increase in pressure will cause the equilibrium to shift toward the formation of N₂O₄ (the reverse reaction), leading to a decrease in the amount of NO₂ present in the system as the number of gas particles decreases.
\[ \text{N}_2\text{O}_4(g) + \text{energy} \rightleftharpoons 2\text{NO}_2(g) \]
we have one mole of dinitrogen tetroxide (N₂O₄) on the left side and two moles of nitrogen dioxide (NO₂) on the right side. When we talk about increasing pressure in a gaseous equilibrium, it's essential to apply Le Chatelier's Principle, which states that if a system at equilibrium experiences a change in pressure, temperature, or concentration, the system will adjust to counteract that change.
When pressure is increased in a gaseous system, the equilibrium will shift in the direction that produces fewer moles of gas. Since the left side (reactants) has 1 mole of N₂O₄ and the right side (products) has 2 moles of NO₂, increasing the pressure will favor the formation of N₂O₄ because it results in fewer total moles of gas in the system.
To break it down:
1. **Effect of Increased Pressure**: When the pressure is increased, the system will favor the side with fewer gas moles to reduce that pressure.
2. **Shifting to the Reverse Reaction**: In this case, the reaction will shift to the left (the reverse reaction) to produce more N₂O₄, which decreases the total number of gas molecules (from 2 moles of NO₂ to 1 mole of N₂O₄).
3. **Rate of Reverse Reaction**: The increase in pressure effectively increases the rate at which the reverse reaction occurs because the equilibrium position is shifting to the left, which means the system is producing more N₂O₄ from NO₂.
In summary, the increase in pressure will cause the equilibrium to shift toward the formation of N₂O₄ (the reverse reaction), leading to a decrease in the amount of NO₂ present in the system as the number of gas particles decreases.
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