Chapter 16: Problem 104
Consider the simple one-step reaction: $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)$$ Since the reaction occurs in a single step, the forward reaction has a rate of \(k_{\text { fort }}[A]\) and the reverse reaction has a rate of \(k_{\text { rate }}[\mathrm{B}] .\) What happens to the rate of the forward reaction when we increase the concentration of A? How does this explain the reason behind Le Chatelier's principle?
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