Answer
$$K_P = 3.3$$
Work Step by Step
1. Calculate the initial pressure for the phosgene gas:
$$P = \frac{nRT}{V} = \frac{(3.00 \times 10^{-2})(0.0821)(800)}{1.50} = 1.31 \space atm$$
2. At equilibrium, this are the partial pressures:
$P_{COCl_2} = 1.31 - x$
$P_{CO} = 0 + x$
$P_{Cl_2} = 0 + x$
3. According to the problem: $P_{CO} = 0.497 \space atm$
$P_{CO} = x = P_{Cl_2} = 0.497 \space atm$
$P_{COCl_2} = 1.31 - 0.497 = 0.81 \space atm$
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:
$$K_P = \frac{( 0.81 )}{( 0.497 )( 0.497 )} = 3.3$$
