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(a) $\begin{array}{cccc}{\text{C}}_{2}{\text{H}}_{2}(g)+\hfill & 2{\text{Br}}_{2}(g)\hfill & \rightleftharpoons & {\text{C}}_{2}{\text{H}}_{2}{\text{Br}}_{4}(g)\hfill \\ x\hfill & \_\_\_\_\_\hfill & & \_\_\_\_\_\hfill \end{array}$
(b) $\begin{array}{cccc}{\text{I}}_{2}(aq)+\hfill & {\text{I}}^{\text{\u2212}}(aq)\hfill & \rightleftharpoons \hfill & {\text{I}}_{3}{}^{\text{\u2212}}(aq)\hfill \\ \_\_\_\_\_\hfill & \_\_\_\_\_\hfill & & x\hfill \end{array}$
(c) $\begin{array}{ccccc}{\text{C}}_{3}{\text{H}}_{8}(g)+\hfill & 5{\text{O}}_{2}(g)\hfill & \rightleftharpoons \hfill & 3{\text{CO}}_{2}(g)+\hfill & 4{\text{H}}_{2}\text{O}(g)\hfill \\ x\hfill & \_\_\_\_\_\hfill & & \_\_\_\_\_\hfill & \_\_\_\_\_\hfill \end{array}$
(b) $\begin{array}{cccc}{\text{I}}_{2}(aq)+\hfill & {\text{I}}^{\text{\u2212}}(aq)\hfill & \rightleftharpoons \hfill & {\text{I}}_{3}{}^{\text{\u2212}}(aq)\hfill \\ -x\hfill & -x\hfill & & x\hfill \end{array}$
(c) $\begin{array}{lllll}{\text{C}}_{3}{\text{H}}_{8}(g)+\hfill & 5{\text{O}}_{2}(g)\hfill & \rightleftharpoons \hfill & 3{\text{CO}}_{2}(g)+\hfill & 4{\text{H}}_{2}\text{O}(g)\hfill \\ x\hfill & 5x\hfill & & \mathrm{-3}x\hfill & \mathrm{-4}x\hfill \end{array}$
(a) $\begin{array}{llll}2{\text{SO}}_{2}(g)+\hfill & {\text{O}}_{2}(g)\hfill & \rightleftharpoons \hfill & 2{\text{SO}}_{3}(g)\hfill \\ \_\_\_\_\_\hfill & x\hfill & & \_\_\_\_\_\hfill \end{array}$
(b) $\begin{array}{lll}{\text{C}}_{4}{\text{H}}_{8}(g)\hfill & \rightleftharpoons \hfill & 2{\text{C}}_{2}{\text{H}}_{4}(g)\hfill \\ \_\_\_\_\_\hfill & & \mathrm{-2}x\hfill \end{array}$
(c) $\begin{array}{lllll}4{\text{NH}}_{3}(g)+\hfill & 7{\text{H}}_{2}\text{O}(g)\hfill & \rightleftharpoons \hfill & 4{\text{NO}}_{2}(g)+\hfill & 6{\text{H}}_{2}\text{O}(g)\hfill \\ \\ \_\_\_\_\_\hfill & \_\_\_\_\_\hfill & & \_\_\_\_\_\hfill & \_\_\_\_\_\hfill \end{array}$
(a) 2 x , x , −2 x; (b) x , −2 x; (c) 4 x , 7 x , −4 x , −6 x or −4 x , −7 x , 4 x , 6 x
Because the value of the reaction quotient of any reaction at equilibrium is equal to its equilibrium constant, we can use the mathematical expression for Q _{c} (i.e., the law of mass action ) to determine a number of quantities associated with a reaction at equilibrium. It may help if we keep in mind that Q _{c} = K _{c} (at equilibrium) in all of these situations and that there are only three basic types of calculations:
A similar list could be generated using Q _{P} , K _{P} , and partial pressure. We will look at solving each of these cases in sequence.
Since the law of mass action is the only equation we have to describe the relationship between K _{c} and the concentrations of reactants and products, any problem that requires us to solve for K _{c} must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for K _{c} , as it will be the only unknown.
[link] showed us how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE chart—for I nitial, C hange, and E quilibrium–will be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and we consider no shift toward equilibrium to have happened. The next row of data is the change that occurs as the system shifts toward equilibrium—do not forget to consider the reaction stoichiometry as described in a previous section of this chapter. The last row contains the concentrations once equilibrium has been reached.
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