Welcome to the article For the following reaction K 0 538 at 313 K text N 2 text O 4 g rightleftharpoons 2 text NO 2 g If a. On this page, you will learn the essential and logical steps to better understand the topic being discussed. We hope the information provided helps you gain valuable insights and is easy to follow. Let’s begin the discussion!
Answer :
Final answer:
The equilibrium concentrations for the given chemical reaction at 313 K are calculated using the equilibrium constant and initial concentration. After solving the resulting quadratic equation, the equilibrium concentrations of N2O4 and NO2 at 313 K are found to be 0.016 mol L−1 and 0.068 mol L−1, respectively.
Explanation:
The given chemical equation can be written as N2O4(g)⇌2 NO2(g), which suggests that for every mole of N2O4 that decomposes, 2 moles of NO2 are produced.
Let's represent the initial concentration of N2O4 as [N2O4]i = 0.0500 mol L–1 and let's denote the change in concentration as x.
At equilibrium, the concentrations of N2O4 and NO2 will be [N2O4] = 0.0500 - x and [NO2] = 2x, respectively.
We can then use the equilibrium constant expression K = [NO2]^2/[N2O4], substituting the values we have.
Given K = 0.538 at 313 K, we set up the equation 0.538 = (2x)^2 / (0.0500 - x).
Solving this quadratic equation for x, we found x = 0.0340 mol L−1.
Substituting x into the expressions for [NO2] and [N2O4], we get [NO2] = 2x = 0.0680 mol L−1 and [N2O4] = 0.0500 - x = 0.016 mol L−1 at equilibrium.
Therefore, the equilibrium concentrations of N2O4 and NO2 at 313K are 0.016 mol L−1 and 0.068 mol L−1, respectively.
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