Question

The \frac{{{K_P}}}{{{K_C}}} ratio will be highest in case of CO(g) + \frac{1}{2}O2(g) ⇌ CO2(g) H2(g) + I2(g) ⇌ 2HI(g) PCl5(g) ⇌ PCl3(g) + Cl2(g) 7H2(g) + 2NO2(g) ⇌ 2NH3(g) + 4H2O(g)

• A. CO(g) + \frac{1}{2}O2(g) ⇌ CO2(g)
• B. H2(g) + I2(g) ⇌ 2HI(g)
• C. PCl5(g) ⇌ PCl3(g) + Cl2(g)
• D. 7H2(g) + 2NO2(g) ⇌ 2NH3(g) + 4H2O(g)

Answer 1

Answer: (C)

PCl5(g) ⇌ PCl3(g) + Cl2(g)

Answer 2

The relationship between $K_P$ and $K_C$ is given by the equation:

$K_P = K_C (RT)^{\Delta n_g}$

where:

$K_P$ is the equilibrium constant in terms of partial pressures. $K_C$ is the equilibrium constant in terms of molar concentrations. $R$ is the ideal gas constant. $T$ is the absolute temperature. $\Delta n_g$ is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.

We need to find the ratio $\frac{K_P}{K_C}$:

$\frac{K_P}{K_C} = (RT)^{\Delta n_g}$

To maximize this ratio, we need to maximize the value of $\Delta n_g$.

Let's calculate $\Delta n_g$ for each given reaction:

1. CO(g) + $\frac{1}{2}$O$_2$(g) $\rightleftharpoons$ CO$_2$(g)

   $\Delta n_g = 1 - (1 + 0.5) = -0.5$

2. H$_2$(g) + I$_2$(g) $\rightleftharpoons$ 2HI(g)

   $\Delta n_g = 2 - (1 + 1) = 0$

3. PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g)

   $\Delta n_g = (1 + 1) - 1 = 1$

4. 7H$_2$(g) + 2NO$_2$(g) $\rightleftharpoons$ 2NH$_3$(g) + 4H$_2$O(g)

   $\Delta n_g = (2 + 4) - (7 + 2) = -3$

The highest value of $\Delta n_g$ is 1, which corresponds to the reaction: PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g)