Answer
\begin{vmatrix}
& [A] (M)& [B] (M) \\
a. & 0.20 & 0.80 \\
b. & 0.33 & 0.67 \\
c. & 0.38 & 1.2
\end{vmatrix}
Work Step by Step
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$
