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

Concept: For gaseous equilibria, \[\begin{aligned} K_p=K_c(RT)^{\Delta n} \end{aligned}\] where \[\begin{aligned} \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} \end{aligned}\] For the reverse reaction, \[\begin{aligned} K'=\frac{1}{K} \end{aligned}\]

Step 1: Calculate \(\Delta n\) for the forward reaction. \[\begin{aligned} H_2+I_2 \rightleftharpoons 2HI \end{aligned}\] \[\begin{aligned} \Delta n=2-(1+1)=0 \end{aligned}\] Hence, \[\begin{aligned} K_p=K_c(RT)^0=K_c \end{aligned}\] Therefore, option (A) is correct.

Step 2: Analyze the reverse reaction. For the reverse reaction, \[\begin{aligned} K'_c=\frac{1}{K_c} \end{aligned}\] and \[\begin{aligned} K'_p=\frac{1}{K_p} \end{aligned}\] Thus, option (D) is correct.

Step 3: Compare \(K'_c\) and \(K'_p\). Since \[\begin{aligned} K_c=K_p \end{aligned}\] their reciprocals are also equal. \[\begin{aligned} K'_c=K'_p \end{aligned}\] Therefore, option (C) is correct.

Step 4: Identify the incorrect relationship. \[\begin{aligned} K'_c=\frac{1}{K_c} \neq K_c \end{aligned}\] Hence, \[\begin{aligned} \boxed{K_c=K'_c} \end{aligned}\] is incorrect. Therefore, option \(\mathbf{(B)}\) is correct.