Answer
$ P_{ NO } = 8.0 \times 10^{-5} \space atm$
$ P_{ O_2 } = 4.0 \times 10^{-5} \space atm$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& P_{ NO_2 }& P_{ NO }& P_{ O_2 }\\
Initial& 0.75 & 0 & 0 \\
Change& -2 x& 2 x& +x\\
Equilibrium& 0.75 -2 x& 0 2 x& 0 +x\\
\end{vmatrix}$$
- The exponent of each pressure is equal to its balance coefficient.
$$K_P = \frac{P_{Products}}{P_{Reactants}} = \frac{P_{ NO } ^{ 2 }P_{ O_2 }}{P_{ NO_2 } ^{ 2 }}$$
2. At equilibrium, these are the concentrations of each compound:
$ P_{ NO_2 } = 0.75 \space M - 2x$
$ P_{ NO } = 2x$
$ P_{ O_2 } = x$
$$4.48 \times 10^{-13} = \frac{(2x)^2( x)}{(0.75-2x)^2}$$
3. Solve for x:
$x = 4.0 \times 10^{-5}$
$ P_{ NO } = 2( 4.0 \times 10^{-5}) = 8.0 \times 10^{-5}$
$ P_{ O_2 } = 4.0 \times 10^{-5}$
