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The general second order equation is $$\frac{\mathrm d[A]}{\mathrm dt}=-k[\ce{A}][\ce{B}]\tag{1}$$ with $[\ce{A}]_0,$ $[\ce{B}]_0$ the initial amounts. At a time $t$ there are $[\ce{A}]_0-x$, and $[\ce{B}]_0-x$ remaining if amount $x$ reacts. Thus $$\frac{\mathrm dx}{\mathrm dt}=k([\ce{A}]_0-x)([\ce{B}]_0-x).\tag{2}$$ You should be able to integrate this ...


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[OP] The equilibrium constant would not include the solid $\ce{I2}$, but why is this? Let me explain this with a different example. If you have a saturated solution (e.g. lemonade with too much sugar in it) it is at equilibrium. If you add more sugar, the lemonade does not get sweeter. That tells you that the amount of solid does not matter (as long as ...


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Here's one way you could get from eqn. (2) to eqn. (3) in porphyrin's answer using Mathematica. I made extensive use of the Part function, whose shorthand is "[[ ]]", to pull out the desired subexpression from each answer. E.g., Part[x+y, 1], which means "take the first part of the expression (x+y)", can be expressed as (x+y)[[1]], and ...


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A reaction's rate does depend upon the consumption of its reactants, and the manner in which the reactants interact. Let's consider the following reaction: $$\ce{A + B -> final products}$$ The rate of consumption of reactants $$\dfrac{-d[A]}{dt}= k[A][B]$$ and $$\dfrac{-d[B]}{dt}= k[A][B]$$ It seems that this reaction is a second order reaction since the ...


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