For the equilibrium: A(g) ⇌ nB(g), the equilibrium constant, , is related with the degree of dissoci — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Define initial moles and set up the equilibrium table For the general gas-phase dissociation equilibrium: \[\ce{A(g) <=> nB(g)}\] Let us assume the initial number of moles of reactant gas \ce{A} is $1\text{ mol}$ and the initial number of moles of product gas \ce{B} is $0\text{ mol}$. Let $\alpha$ represent the degree of dissociation of reactant gas \ce{A} at equilibrium. The stoichiometry of the reaction dictates that for every $1\text{ mol}$ of \ce{A} that dissociates, $n\text{ mol}$ of \ce{B} are produced. We can construct an equilibrium table for the system: \[\begin{array}{lccc} \text{Species} & \ce{A(g)} & \ce{<=>} & \ce{nB(g)} \\ \hline \text{Initial moles } (t = 0) & 1 & & 0 \\ \text{Change in moles} & -\alpha & & +n\alpha \\ \text{Equilibrium moles} & 1 - \alpha & & n\alpha \\ \hline \end{array}\] Step 2 - Calculate the total number of gaseous moles at equilibrium The total number of moles of all gaseous species present in the mixture at equilibrium ($n_{\text{total}}$) is: \[n_{\text{total}} = n_{\ce{A}} + n_{\ce{B}}\] \[n_{\text{total}} = (1 - \alpha) + n\alpha\] By factoring out $\alpha$, we can simplify this expression: \[n_{\text{total}} = 1 + (n - 1)\alpha\] Step 3 - Express the partial pressure of each gaseous component According to Dalton's Law of Partial Pressures, the partial pressure ($p_i$) of a component in a gas mixture is the product of its mole fraction ($x_i$) and the total equilibrium pressure ($P$): \[p_i = \left(\frac{n_i}{n_{\text{total}}}\right) P\] Applying this definition to our equilibrium species: * **Partial pressure of reactant gas \ce{A} ($p_{\ce{A}}$):** \[p_{\ce{A}} = \frac{1 - \alpha}{1 + (n - 1)\alpha} \cdot P\] * **Partial pressure of product gas \ce{B} ($p_{\ce{B}}$):** \[p_{\ce{B}} = \frac{n\alpha}{1 + (n - 1)\alpha} \cdot P\] Step 4 - Substitute the partial pressures into the $K_p$ expression The equilibrium constant in terms of partial pressures ($K_p$) is defined as: \[K_p = \frac{(p_{\ce{B}})^n}{p_{\ce{A}}}\] Substituting the expressions for $p_{\ce{A}}$ and $p_{\ce{B}}$: \[K_p = \frac{\left[ \frac{n\alpha}{1 + (n - 1)\alpha} \cdot P \right]^n}{\frac{1 - \alpha}{1 + (n - 1)\alpha} \cdot P}\] Step 5 - Simplify the algebraic expression Now, let us separate the terms to simplify the fraction: \[K_p = \frac{(n\alpha)^n \cdot P^n}{\left[1 + (n - 1)\alpha\right]^n} \times \frac{1 + (n - 1)\alpha}{(1 - \alpha) \cdot P}\] Combine the total pressure terms ($P$): \[\frac{P^n}{P} = P^{n-1}\] Combine the terms involving the factor $[1 + (n - 1)\alpha]$: \[\frac{1 + (n - 1)\alpha}{[1 + (n - 1)\alpha]^n} = \frac{1}{[1 + (n - 1)\alpha]^{n-1}}\] Substituting these simplified terms back into the overall expression for $K_p$ yields: \[K_p = \frac{(n\alpha)^n \cdot P^{n-1}}{(1 - \alpha) \cdot [1 + (n - 1)\alpha]^{n-1}}\] Thus, the equilibrium constant $K_p$ is: \[K_p = \boxed{\frac{(n\alpha)^n \cdot P^{n-1}}{(1 - \alpha) \cdot [1 + (n - 1)\alpha]^{n-1}}}\] Step 6 - Evaluate the options * **Option (A)**: Correct. This is the exact algebraic relationship derived using Dalton's law of partial pressures and the definition of the equilibrium constant $K_p$. * **Option (B)**: Incorrect. This option has a power of $n$ instead of $n-1$ in the denominator term $[1 + (n - 1)\alpha]^n$. * **Option (C)**: Incorrect. This option has a power of $n-1$ instead of $n$ for the term $(n\alpha)$ in the numerator. * **Option (D)**: Incorrect. This option lacks the power of $n$ for the term $(n\alpha)$ in the numerator.