Consider the equilibrium for the reaction N2O4 (g) = 2 NO2 (g) delta H=58 KJ. Indicated in which direction the equilibrium will shift under the following conditions. Explain your answer:
a. N2O4 is added
b. NO2 is removed
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This problem is about the equilibrium of the gas at a certain temperature. The equation
`N2O4 (g) <=> 2 NO2 (g)`
means that for every mole of N2O4, 2 moles of NO2 are produced and vice versa until it reaches the point where the rate of production of N2O4 is equal with the rate of production of NO2.
If we disrupt the equilibrium by (a.) adding N2O4, the system will find a way to equilibrate the system by producing more NO2. Therefore the shift will be from left to right (shift to the right). Added N2O4 will be consumed to produce NO2 until it reaches equilibrium again.
If we (b.) remove NO2, the system will make new NO2 out of the remaining N2O4 in the container. Therefore the shift will be from left to right favoring the formation of new NO2.
When being asked a question about the direction of equilibrium under certain conditions, you can refer to Le Chatelier's principle. This principle basically states that when something happens in the reaction/system, equilibrium will shift to regain balance in the reaction.
In this reaction you have N2O4 <--> 2NO2
When N2O4 is added, you're adding more reactant, making the left side heavier. The equilibrium will shift RIGHT to make more products to regain balance and achieve equilibrium again. Also remember, the more the reactants, the more the products because you have more "stuff" to make products with.
When NO2 is removed, you're taking away product, making the left side heavier. In this situation equilibrium will shift RIGHT to make more products again to achieve equilibrium again. You lost product but you don't want to lose it. You want to regain. So you'll make more product and that's why equilibrium shifts right.
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