Question

The ratio \( \frac{K_p}{K_c} \) for the reaction \[ {CO}(g) + \frac{1}{2} {O}_2(g) \rightleftharpoons {CO}_2(g) \] is:

• A. \( (RT)^{1/2} \)
• B. \( RT \)
• C. 1
• D. \( \frac{1}{\sqrt{RT}} \)

Hint:

The ratio \( \frac{K_p}{K_c} \) depends on the change in the number of moles of gas between the products and reactants, and it is governed by the equation \( K_p = K_c (RT)^{\Delta n} \).

Answer 1

The Correct Option is D

Step 1: {Relation Between \( K_p \) and \( K_c \)} 
The equilibrium constant \( K_p \) and \( K_c \) are related by the following equation: \[ K_p = K_c (RT)^{\Delta n} \] where \( \Delta n \) is the change in the number of moles of gas between products and reactants. 
Step 2: {Calculate \( \Delta n \)} 
For the given reaction: \[ \Delta n = {moles of products} - {moles of reactants} = 1 - \left( 1 + \frac{1}{2} \right) = -\frac{1}{2} \] 
Step 3: {Substitute \( \Delta n \)} 
Substitute \( \Delta n = -\frac{1}{2} \) into the relation: \[ K_p = K_c (RT)^{-\frac{1}{2}} \] Thus, the correct answer is (D). 
 

Answer 2

Step 1: Use the relationship between \( K_p \) and \( K_c \)
The general formula connecting \( K_p \) and \( K_c \) is:
\[ K_p = K_c(RT)^{\Delta n} \]
Where:
- \( R \) = gas constant
- \( T \) = temperature in Kelvin
- \( \Delta n \) = moles of gaseous products − moles of gaseous reactants

Step 2: Analyze the reaction
\[ CO(g) + \frac{1}{2}O_2(g) \rightleftharpoons CO_2(g) \]
Moles of gaseous reactants = \( 1 + \frac{1}{2} = \frac{3}{2} \)
Moles of gaseous products = 1
\[ \Delta n = 1 - \frac{3}{2} = -\frac{1}{2} \]

Step 3: Apply the formula
\[ K_p = K_c(RT)^{\Delta n} = K_c(RT)^{-1/2} \]
\[ \Rightarrow \frac{K_p}{K_c} = (RT)^{-1/2} = \frac{1}{\sqrt{RT}} \]

Final Answer: \( \frac{1}{\sqrt{RT}} \)