## Chemistry: Molecular Approach (4th Edition)

a. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ A ]& [ B ]\\ Initial& 1.0 & 0 \\ Change& -x& +x\\ Equilibrium& 1.0 -x& 0 +x\\ \end{vmatrix}$$ 2. At equilibrium, these are the concentrations of each compound: $[ A ] = 1.0 \space M - x$ $[ B ] = 0 \space M + x$ 3. The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ B ]}{[ A ]}$$ $$4.0 = \frac{( x)}{(1.0 - x)}$$ $$(4.0)(1.0) - 4.0x = x$$ $$x = \frac{4.0}{5.0} = 0.80 \space M$$ $[ A ] = 1.0 \space M - 0.80 \space M = 0.20 \space M$ $[ B ] = 0.80 \space M$ b. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ A ]& [ B ]\\ Initial& 1.0 & 0 \\ Change& -2 x& 2 x\\ Equilibrium& 1.0 -2 x& 0 + 2 x\\ \end{vmatrix}$$ 2. At equilibrium, these are the concentrations of each compound: $[ A ] = 1.0 \space M - 2x$ $[ B ] = 0 \space M + 2x$ 3. The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ B ] ^{ 2 }}{[ A ] ^{ 2 }}$$ $$4.0 = \frac{(2x)^2}{(1.0-2x)^2}$$ $x = 0.33\underline{333}$ or $x = 1.0$ But, if x = 1.0, then: $[A] = 1.0 - 2.0 = -1.0$, which is invalid, because concentration can't be negative. $x =0.33\underline{333}$ $[ A ] = 1.0 - 2(0.33\underline{333}) = 0.33 \space M$ $[ B ] = 2(0.33\underline{333}) = 0.67 \space M$ c. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ A ]& [ B ]\\ Initial& 1.0 & 0 \\ Change& -x& 2 x\\ Equilibrium& 1.0 -x& 0 + 2 x\\ \end{vmatrix}$$ 2. At equilibrium, these are the concentrations of each compound: $[ A ] = 1.0 \space M - x$ $[ B ] = 0 \space M + 2x$ 3. The exponent of each concentration is equal to its balance coefficient. $$K_C = \frac{[Products]}{[Reactants]} = \frac{[ B ] ^{ 2 }}{[ A ]}$$ $$4.0 = \frac{(2x)^2}{(1.0 - x)}$$ $x = 0.62$ (Only positive root) $[ A ] = 1.0 \space M - 0.62 \space M = 0.38 \space M$ $[ B ] = 2(0.62) = 1.2 \space M$