At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops? - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-08-20T08:21:36Z https://chemistry.stackexchange.com/feeds/question/4475 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://chemistry.stackexchange.com/q/4475 2 At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops? sanchitkhanna26 https://chemistry.stackexchange.com/users/1400 2013-03-24T11:59:08Z 2016-01-25T15:57:35Z <p>This one is troubling me, and I even got it wrong in my exam:</p> <blockquote> <p>at equilibrium the mass of reactants and products are equal, does that mean the reaction stops?</p> </blockquote> <p>Please explain this one. </p> https://chemistry.stackexchange.com/questions/4475/-/4479#4479 3 Answer by ManishEarth for At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops? ManishEarth https://chemistry.stackexchange.com/users/22 2013-03-25T08:25:13Z 2013-03-25T08:25:13Z <p>At equilibrium, the reaction continues, but the rate of the forward reaction is equal to the rate of the backward reaction.</p> <p>Let's consider the first order reaction $\ce{A &lt;=&gt;[k_f][k_b] B}$</p> <p>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.</p> <p>Here's a diagram for how the rates of the reaction proceed for the formation of ammonia. Note that the rates never reach 0:</p> <p><img src="https://i.stack.imgur.com/PpHJq.png" alt="enter image description here"></p> https://chemistry.stackexchange.com/questions/4475/-/44204#44204 -1 Answer by Anas Qaisrani for At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops? Anas Qaisrani https://chemistry.stackexchange.com/users/25045 2016-01-25T15:26:22Z 2016-01-25T15:26:22Z <p>Because at equilibrium state forward and reverse reactions occur at same place but they occur in opposite directions so reaction don't stops..</p> https://chemistry.stackexchange.com/questions/4475/-/44209#44209 1 Answer by Jan for At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops? Jan https://chemistry.stackexchange.com/users/7475 2016-01-25T15:57:35Z 2016-01-25T15:57:35Z <p>A reaction at equilibrium never stops (by itself), it rather remains in a <em>dynamic</em> equilibrium where forwards and reverse reactions occur <em>at the same rate.</em> However, macroscopically you cannot differentiate between a reaction that stopped and a dynamic equilibrium.</p> <p>Also another note: It is <strong>wrong</strong> to say that the masses of reactants and products be equal at equilibrium. You cannot even say that the amounts, concentrations or any other variable is equal for reactants and products. Rather, for each reaction there is a specific ratio of concentrations that is reached at equilibrium, commonly known as the equilibrium constant.</p> <p>$$\ce{aA + bB &lt;=&gt; cC + dD}$$</p> <p>$$k_\mathrm{eq} = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$</p> <p>The mass ratio at an equilibrium can be anything. Most notably, it can be both $1\,:\,1$ and $10^{99}\,:\,1$ and there is no rule as to which mass ratio would be present at a given equilibrium.</p>