Answer
$$K_c = 0.0353 $$
Work Step by Step
1. Calculate the initial pressure for the phosgene gas:
$$P = \frac{nRT}{V} = \frac{(2.50)(0.0821)(400+273.15)}{1.50} = 92.1 \space atm$$
2. At equilibrium, this are the partial pressures:
$P_{NOCl} = 92.1 - 2x$
$P_{NO} = 0 + 2x$
$P_{Cl_2} = 0 + x$
3. According to the problem: $0.28 = \frac{2x}{92.1}$. Solving for x:
$0.28 \times 92.1 = 2x$
$$x = \frac{0.28 \times 92.1}{2} = 12.9$$
4. Substituting:
$P_{NOCl} =66.3 \space atm$
$P_{NO} = 25.8 \space atm$
$P_{Cl_2} = 12.9 \space atm$
4. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_P = \frac{[Products]}{[Reactants]} = \frac{P_{NO}^2P_{Cl_2}}{P_{ NOCl}^2}$$
5. Substitute the values and calculate the constant value:
$$K_P = \frac{25.8^2 \times 12.9}{66.3^2} = 1.95$$
6. Calculate $\Delta n$ (n is the amount of moles of gases):
$$\Delta n = n_{products} - n_{reactants} = 3 - 2 = 1 $$
7. Convert the temperature to Kelvin:
$$T/K = 400 + 273.15 = 673.15 $$
8. Calculate Kc:
$$K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{( 1.95 )}{(0.0821 \times 673.15 )^{ 1 }}$$
$$K_c = 0.0353 $$
