Answer
$$K_C = 6.01 \times 10^{-6}$$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ NH_3 ]& [ O_2 ]& [ N_2 ]& [ H_2O ]\\
Initial& 0.0150 & 0.0150 & 0 & 0 \\
Change& -4 x& -3 x& 2 x& 6 x\\
Equilibrium& 0.0150 -4 x& 0.0150 -3 x& 2 x& 6 x\\
\end{vmatrix}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ NH_3 ] = 0.0150 \space M - 4x$
$ [ O_2 ] = 0.0150 \space M - 3x$
$ [ N_2 ] = 0 \space M + 2x$
$ [ H_2O ] = 0 \space M + 6x$
3. Using the concentration of $ N_2 $ at equilibrium, find x:
$ 2 x = 1.96 x 10^{-3} $
$ x = \frac{ 1.96 x 10^{-3} }{ 2 } $
$x = 9.80 \times 10^{-4} $
$ [ NH_3 ] = 0.0150 \space M - 4*( 9.80 \times 10^{-4} ) = 0.01108 $
$ [ O_2 ] = 0.0150 \space M - 3*( 9.80 \times 10^{-4} ) = 0.01206 $
$ [ N_2 ] = 2*( 9.80 \times 10^{-4} )=1.96 \times 10^{-3} $
$ [ H_2O ] = 6*( 9.80 \times 10^{-4} )=5.88 \times 10^{-3} $
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ N_2 ] ^{ 2 }[ H_2O ] ^{ 6 }}{[ NH_3 ] ^{ 4 }[ O_2 ] ^{ 3 }}$$
4. Substitute the values and calculate the constant value:
$$K_C = \frac{( 1.96 \times 10^{-3} )^{ 2 }( 5.88 \times 10^{-3} )^{ 6 }}{( 0.01108 )^{ 4 }( 0.01206 )^{ 3 }} = 6.01 \times 10^{-6}$$
