19.1: Reversible Reaction

Last updated
Save as PDF

Page ID: 53903

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

A solution of cobalt chloride in water is pink due to the presence of the solvated \(\ce{Co^{2+}}\) ion. If sufficient \(\ce{HCl}\) is added, the solution turns blue as the \(\ce{CoCl_4^{2-}}\) ion forms. The reaction can be shifted back to the pink form if more water is added to the solution.

Chemical Reversibility

Up until this point, we have written the equations for chemical reactions in a way that would seem to indicate that all reactions proceed completely until all the reactants have been converted into products. In reality, a great many chemical reactions do not proceed entirely to completion. A reversible reaction is a reaction in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously. One example of a reversible reaction is the reaction of hydrogen gas and iodine vapor to form hydrogen iodide. The forward and reverse reactions can be written as follows.

\[\text{Forward reaction:} \: \: \: \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)\nonumber \]

\[\text{Reverse reaction:} \: \: \: 2 \ce{HI} \left( g \right) \rightarrow \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right)\nonumber \]

In the forward reaction, hydrogen and iodine combine to form hydrogen iodide. In the reverse reaction, hydrogen iodide decomposes back into hydrogen and iodine. The two reactions can be combined into one equation by the use of a double arrow:

\[\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightleftharpoons 2 \ce{HI} \left( g \right)\nonumber \]

The double arrow is the indication that the reaction is reversible.

When hydrogen and iodine gases are mixed in a sealed container, they begin to react and form hydrogen iodide. At first, only the forward reaction occurs because no \(\ce{HI}\) is present. As the forward reaction proceeds, it begins to slow down as the concentrations of the \(\ce{H_2}\) and the \(\ce{I_2}\) decrease. As soon as some \(\ce{HI}\) has formed, it begins to decompose back into \(\ce{H_2}\) and \(\ce{I_2}\). The rate of the reverse reaction starts out slow because the concentration of \(\ce{HI}\) is low. Gradually, the rate of the forward reaction decreases while the rate of the reverse reaction increases. Eventually the rate of combination of \(\ce{H_2}\) and \(\ce{I_2}\) to produce \(\ce{HI}\) becomes equal to the rate of decomposition of \(\ce{HI}\) into \(\ce{H_2}\) and \(\ce{I_2}\). When the rates of the forward and reverse reactions have become equal to one another, the reaction has achieved a state of balance.