Question:medium

\(\dfrac{K_p}{K_c}\) for the reaction at \(T(K)\) is \[ CO(g)+\frac{1}{2}O_2(g)\rightleftharpoons CO_2(g) \]

Show Hint

Always use \[ K_p=K_c(RT)^{\Delta n} \] where \[ \Delta n=(\text{gaseous product moles})-(\text{gaseous reactant moles}) \] Only gaseous species are considered while calculating \(\Delta n\).
Updated On: Jul 18, 2026
  • \(\sqrt{RT}\)
  • \(2RT\)
  • \(RT\)
  • \(\dfrac{1}{\sqrt{RT}}\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Count gaseous moles on each side.
\[ CO(g)+\tfrac{1}{2}O_2(g)\rightleftharpoons CO_2(g) \]
The reactant side has $1+\tfrac{1}{2}=\tfrac{3}{2}$ moles of gas, the product side has $1$ mole of gas.

Step 2: Find $\Delta n_g$.
\[ \Delta n_g=1-\tfrac{3}{2}=-\tfrac{1}{2} \]

Step 3: Apply $K_p=K_c(RT)^{\Delta n_g}$.
\[ \frac{K_p}{K_c}=(RT)^{-1/2} \]

Step 4: Rewrite the negative half power as a root.
\[ (RT)^{-1/2}=\frac{1}{\sqrt{RT}} \]
\[ \boxed{\frac{1}{\sqrt{RT}}} \]
matching option (4).
Was this answer helpful?
0