Answer
$K_p$ for this reaction at 800 K is equal to 5.20
Work Step by Step
1.
$P_{CO} = P_{Cl_2} = 0.497$ atm
2.
$[COCl_2] = (3.00 \times 10^{-2} mol)/1.50 L = 0.0200 \space M$
3.
$P_{COCl_2} = (0.0200)(0.08206)(800) = 1.31$ atm
4. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_P = \frac{[Products]}{[Reactants]} = \frac{P_{ COCl_2 }}{P_{ CO }P_{ Cl_2 }}$$
5. Substitute the values and calculate the constant value:
$$K_P = \frac{( 1.31 )}{( 0.497 )( 0.497 )} = 5.20$$
