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Chemistry · Ch 12 — Chemical Equilibrium

Law of Mass Action

12.4.2

Law of Mass Action

The Law of Mass Action states that the rate of a chemical reaction at each instant is proportional to the product of the concentration terms of all the reactants, with each concentration raised to a power equal to the number of molecules of that reactant appearing in the balanced chemical equation. For a general reaction A+B→CA+B\rightarrow C, where A and B are reactants (concentrations expressed in mol L−1^{-1}, written in square brackets) and C is the product, the law of mass action gives the proportionality Rate ∝[A][B]\propto[A][B]. Introducing a proportionality constant kk, called the rate constant of the reaction, turns this into the rate equation, Rate =k[A][B]=k[A][B]. If a reactant's stoichiometric coefficient is greater than 1, its concentration is raised to that power in the rate expression -- Problem 12.1 works through exactly this for C(s)+O2(g)→CO2(g)C(s)+O_2(g)\rightarrow CO_2(g) (Rate =k[C][O2]=k[C][O_2]) and 2KClO3(s)→2KCl(s)+3O2(g)2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g) (Rate =k[KClO3]2=k[KClO_3]^2, since KClO3 has coefficient 2). In any reversible reaction carried out in a closed system, the forward reaction's rate is high at the start (when only reactants are pre …

Misc Problem 12.1Rate equations for combustion and KClO3 decomposition

Worked out. Worked example applying the law of mass action to write rate equations for two reactions. (i) C(s) + O2(g) -> CO2(g): the reactants are C and O2, so Rate is proportional to [C][O2], giving Rate = k[C][O2]. (ii) 2KClO3(s) -> 2KCl(s) + 3O2(g): the single reactant KClO3 has a coefficient of 2 in the balanced equation, so its concentration is raised to the power 2, giving Rate is proportional to [KClO3]^2, i.e. Rate = k[KClO3]^2. This shows explicitly how a stoichiometric coefficient greater than 1 becomes an exponent …