For the reaction, $$CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)$$ the $$\frac{K_p}{K_c}$$ is equal to

The relationship between the equilibrium constants $$K_p$$ (expressed in terms of partial pressures) and $$K_c$$ (expressed in terms of molar concentrations) is given by the formula:

$$K_p = K_c(RT)^{\Delta n_g}$$

Where:

• $$R$$ = Universal gas constant
• $$T$$ = Absolute temperature (in Kelvin)
• $$\Delta n_g$$ = (Number of moles of gaseous products) $$-$$ (Number of moles of gaseous reactants)

For the given chemical equation:

$$CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)$$

• Moles of gaseous products ($$n_p4$): $$1$$ (from $$COCl_2$$)
• Moles of gaseous reactants ($$n_r$$): $$1 + 1 = 2$$ (from $$CO$$ and $$Cl_2$$)

$$\Delta n_g = 1 - 2 = -1$$

Substitute the value of $$\Delta n_g = -1$$ back into the formula:

$$K_p = K_c(RT)^{-1}$$

To find the ratio $$\frac{K_p}{K_c}$$, divide both sides by $$K_c$$:

$$\frac{K_p}{K_c} = (RT)^{-1} = \frac{1}{RT}$$