Condition suitable for forming atomic chlorine from molecular chlorine is: — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Dissociation Reaction and Identify Enthalpy Change The formation of atomic chlorine from molecular chlorine: \[\ce{Cl2(g) <=> 2Cl(g)}\] Breaking a \ce{Cl-Cl} covalent bond requires energy input. The forward reaction is endothermic: \[\Delta H^\circ > 0\] Step 2 - Determine \(\Delta n_g\) \[\Delta n_g = n_{\text{g, products}} - n_{\text{g, reactants}} = 2 - 1 = +1\] Step 3 - Effect of Temperature (Le Chatelier's Principle) * The forward reaction is endothermic ($\Delta H^\circ > 0$). * By Le Chatelier's principle, increasing temperature shifts equilibrium in the endothermic (forward) direction. * $\therefore$ **High temperature** favors formation of atomic chlorine. Low temperature suppresses it. Step 4 - Effect of Pressure (Le Chatelier's Principle) * Since $\Delta n_g = +1$, the forward reaction produces more gaseous moles. * By Le Chatelier's principle, decreasing pressure shifts equilibrium toward more gas molecules (forward direction). * $\therefore$ **Low pressure** favors formation of atomic chlorine. High pressure shifts equilibrium backward. Step 5 - Evaluate Each Option * **(A) Low temperature**: Incorrect. Low temperature suppresses the endothermic forward reaction. * **(B) Low pressure**: Correct. $\Delta n_g = +1$, so low pressure shifts equilibrium forward. * **(C) High temperature**: Correct. High temperature favors the endothermic forward reaction. * **(D) High pressure**: Incorrect. High pressure shifts equilibrium backward toward fewer gas moles. \[\text{Correct Options: } \boxed{\text{B, C}}\]