Question

The equilibrium constants in terms of molar concentration and partial pressure for two reactions are given below. On the basis of these, find the incorrect relationship out of the options given: align* H_2(g)+I_2(g) 2HI(g) K_c and K_p align* align* 2HI(g) H_2(g)+I_2(g) K'_c and K'_p align*

• A. \(K_c=K_p\)
• B. \(K_c=K'_c\)
• C. \(K'_c=K'_p\)
• D. \(K'_p=\dfrac{1}{K_p}\)

Hint:

For reverse reactions, \[ K_{\text{reverse}} = \frac{1}{K_{\text{forward}}} \] and if \(\Delta n=0\), \[ K_p=K_c \] which greatly simplifies equilibrium calculations.

Answer 1

The Correct Option is B

Step 1: Recall the $K_p$ and $K_c$ link.
They are related by $K_p = K_c (RT)^{\Delta n}$, where $\Delta n$ is moles of gas products minus reactants.

Step 2: Check the first reaction.
For $H_2 + I_2 \rightleftharpoons 2HI$, moles go from $2$ to $2$, so $\Delta n = 0$. Then $(RT)^0 = 1$, giving $K_p = K_c$. So that relation is fine.

Step 3: Check the reverse reaction.
For $2HI \rightleftharpoons H_2 + I_2$, again $\Delta n = 0$, so $K'_p = K'_c$ as well. That relation is fine.

Step 4: Relate forward and reverse.
The reverse reaction constant is the reciprocal of the forward one, so $K'_c = \dfrac{1}{K_c}$ and $K'_p = \dfrac{1}{K_p}$. The last option is correct.

Step 5: Test the claim $K_c = K'_c$.
Since $K'_c = \dfrac{1}{K_c}$, saying $K_c = K'_c$ would force $K_c = 1$, which is not generally true. So this relation is wrong.

Step 6: Pick the incorrect one.
The incorrect relationship is $K_c = K'_c$, the second option.
\[ \boxed{K_c = K'_c\ \text{(incorrect)}} \]