Answer
$K_c$ for that reaction at that temperature is equal to 0.12
Work Step by Step
0. Calculate the concentrations.
$$[A] = ( 1.00 )/(0.400) = 2.50 M$$
$$[B] = ( 1.00 )/(0.400) = 2.50 M$$
$$[C] = ( 0.20 )/(0.400) = 0.50 M$$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ C ][ D ] ^{ 2 }}{[ A ][ B ]}$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ A ]& [ B ]& [ C ]& [ D ]\\
Initial& 2.50 & 2.50 & 0 & 0 \\
Change& -x& -x& +x& +2 x\\
Equilibrium& 2.50 -x& 2.50 -x& x& 2 x\\
\end{vmatrix}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ A ] = 2.50 \space M - x$
$ [ B ] = 2.50 \space M - x$
$ [ C ] = 0 \space M + x$
$ [ D ] = 0 \space M + 2x$
3. Using the concentration of $ C $ at equilibrium, find x:
$ 0 + x = 0.50 $
$x = 0.50 $
$ [ A ] = 2.50 \space M - 0.50 =2.0 $
$ [ B ] = 2.50 \space M - 0.50 =2.0 $
$ [ C ] = 0 \space M + 0.50 =0.50 $
$ [ D ] = 0 \space M + 2*( 0.50 )=1.0 $
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ C ][ D ] ^{ 2 }}{[ A ][ B ]}$$
2. Substitute the values and calculate the constant value:
$$K_C = \frac{( 0.50 )( 1.0 )^{ 2 }}{( 2.0 )( 2.0 )} = 0.12$$
