This one is troubling me, and I even got it wrong in my exam:
at equilibrium the mass of reactants and products are equal, does that mean the reaction stops?
Please explain this one.
Answer
At equilibrium, the reaction continues, but the rate of the forward reaction is equal to the rate of the backward reaction.
Let's consider the first order reaction $\ce{A <=>[k_f][k_b] B}$
So, the rate of the forward reaction is $k_f[A]$, and the rate of the backward reaction is $k_b[B]$. Initially, $[B]=0$, and $[A]$ is high, so the forward reaction is very fast, and the backwards reaction doesn't occur at all. As time passes, $[A]$ decreases and $[B]$ increases, so the forward reaction becomes slower and slower, and the backwards reaction speeds up. At one point, both reactions have equal rates ($k_f[A]=k_b[B]$). Here, both reactions take place, but for every time when an $A$ becomes $B$, another $B$ becomes $A$. The reaction is taking place, just that you don;t notice it since the net effect is zero.
Here's a diagram for how the rates of the reaction proceed for the formation of ammonia. Note that the rates never reach 0:
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