Following three gaseous equilibrium reactions are occuring at `27^@C`
A, `2CO+O_2hArr 2CO_2`
B, `PCI_5hArr PCI_3+CI_2`
C, `2HIhArr H_2+I_2`
The correct order of `(K_p)/(K_c)` for the following reactions is

Options

A. `A lt C lt B`
B. `A lt B lt C`
C. `C lt B lt A`
D. `B lt C lt A`

Answer

To solve the problem, we need to find the correct order of \( \frac{K_p}{K_c} \) for the given reactions A, B, and C. We will use the relation between \( K_p \) and \( K_c \):

\[ K_p = K_c \left( RT \right)^{\Delta n} \]

Where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) is the temperature in Kelvin
- \( \Delta n = n_p - n_r \) (the change in the number of moles of gas, where \( n_p \) is the total number of moles of gaseous products and \( n_r \) is the total number of moles of gaseous reactants)

### Step 1: Analyze Reaction A
For reaction A:
\[ 2CO + O_2 \rightleftharpoons 2CO_2 \]

- Products: 2 moles of \( CO_2 \) (thus, \( n_p = 2 \))
- Reactants: 2 moles of \( CO \) + 1 mole of \( O_2 \) (thus, \( n_r = 3 \))

Calculating \( \Delta n \):
\[ \Delta n = n_p - n_r = 2 - 3 = -1 \]

Thus,
\[ \frac{K_p}{K_c} = (RT)^{-1} = \frac{1}{RT} \]

### Step 2: Analyze Reaction B
For reaction B:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

- Products: 1 mole of \( PCl_3 \) + 1 mole of \( Cl_2 \) (thus, \( n_p = 2 \))
- Reactants: 1 mole of \( PCl_5 \) (thus, \( n_r = 1 \))

Calculating \( \Delta n \):
\[ \Delta n = n_p - n_r = 2 - 1 = 1 \]

Thus,
\[ \frac{K_p}{K_c} = (RT)^{1} = RT \]

### Step 3: Analyze Reaction C
For reaction C:
\[ 2HI \rightleftharpoons H_2 + I_2 \]

- Products: 1 mole of \( H_2 \) + 1 mole of \( I_2 \) (thus, \( n_p = 2 \))
- Reactants: 2 moles of \( HI \) (thus, \( n_r = 2 \))

Calculating \( \Delta n \):
\[ \Delta n = n_p - n_r = 2 - 2 = 0 \]

Thus,
\[ \frac{K_p}{K_c} = (RT)^{0} = 1 \]

### Step 4: Compare \( \frac{K_p}{K_c} \) for all reactions
Now we can summarize the results:
- For reaction A: \( \frac{K_p}{K_c} = \frac{1}{RT} \) (less than 1)
- For reaction B: \( \frac{K_p}{K_c} = RT \) (greater than 1)
- For reaction C: \( \frac{K_p}{K_c} = 1 \)

### Step 5: Order of \( \frac{K_p}{K_c} \)
Thus, the order from least to greatest is:
\[ A < C < B \]

### Final Answer
The correct order of \( \frac{K_p}{K_c} \) for the reactions is:
\[ A < C < B \]