For the reaction: (g) ⇌ 2I(g), Kc = 1.0 × 10^-2 mol/L. What volume of the vessel should be taken so — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Express the equilibrium concentrations in terms of vessel volume Let $V$ be the volume of the closed reaction vessel in liters ($\text{L}$). The molar concentration ($[\text{C}]$) of any chemical species is defined as the number of moles ($n$) of that species divided by the total volume of the vessel ($V$): \[[\text{C}] = \frac{n}{V}\] We are given the following quantities at equilibrium: * Number of moles of gaseous iodine molecules ($\ce{I2(g)}$), $n_{\ce{I2}} = 1\text{ mol}$ * Number of moles of gaseous iodine atoms ($\ce{I(g)}$), $n_{\ce{I}} = 0.5\text{ mol}$ Using these values, we write the equilibrium concentrations of the reactants and products: \[[\ce{I2}] = \frac{n_{\ce{I2}}}{V} = \frac{1}{V}\text{ mol L}^{-1}\] \[[\ce{I}] = \frac{n_{\ce{I}}}{V} = \frac{0.5}{V}\text{ mol L}^{-1}\] Step 2 - Write the expression for the equilibrium constant ($K_c$) The balanced chemical equation representing the homogeneous dissociation of iodine gas is: \[\ce{I2(g) <=> 2I(g)}\] The concentration-based equilibrium constant ($K_c$) is expressed according to the law of chemical equilibrium as: \[K_c = \frac{[\ce{I}]^2}{[\ce{I2}]}\] Step 3 - Substitute concentrations and solve for the volume $V$ Substituting the volume-dependent concentrations from Step 1 into the equilibrium constant expression: \[K_c = \frac{\left(\frac{n_{\ce{I}}}{V}\right)^2}{\left(\frac{n_{\ce{I2}}}{V}\right)} = \frac{n_{\ce{I}}^2}{n_{\ce{I2}} \cdot V}\] Now, substitute the given numerical values: * $K_c = 1.0 \times 10^{-2}\text{ mol L}^{-1}$ * $n_{\ce{I2}} = 1\text{ mol}$ * $n_{\ce{I}} = 0.5\text{ mol}$ This substitution yields: \[1.0 \times 10^{-2}\text{ mol L}^{-1} = \frac{(0.5\text{ mol})^2}{1\text{ mol} \times V}\] \[1.0 \times 10^{-2} = \frac{0.25}{V}\] To isolate the volume $V$, rearrange the equation: \[V = \frac{0.25}{1.0 \times 10^{-2}}\text{ L}\] \[V = \frac{0.25}{0.01}\text{ L}\] \[V = \boxed{25\text{ L}}\] Step 4 - Analyze and evaluate the options * **(A) 25 L**: Correct. A vessel volume of 25 L perfectly satisfies the equilibrium constant requirement with the given mole quantities. * **(B) 0.04 L**: Incorrect. This value is obtained from an algebraic error where the equation is inverted as $V = K_c / n_{\ce{I}}^2 = 0.01 / 0.25 = 0.04\text{ L}$. * **(C) 0.25 L**: Incorrect. This value simply represents the numerator term $n_{\ce{I}}^2 = (0.5)^2 = 0.25$ without dividing by $K_c$. * **(D) 5 L**: Incorrect. This is the square root of the correct volume, which has no physical or mathematical basis in this equation.