For the reaction: 2NOCl(g) ⇌ 2(g) + (g), δ H° = 18 kcal and δ S° = 30 cal/K at 300 K. The equilibriu — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Convert the standard enthalpy change to consistent units We are given the standard enthalpy change ($\Delta H^\circ$) in kilocalories ($\text{kcal}$) and the standard entropy change ($\Delta S^\circ$) in calories per Kelvin ($\text{cal K}^{-1}$). To ensure consistency during our thermodynamic calculations, we must first convert $\Delta H^\circ$ from kilocalories to calories: \[\Delta H^\circ = 18\text{ kcal} \times 10^3\text{ cal/kcal} = 18000\text{ cal}\] Step 2 - Calculate the standard Gibbs free energy change ($\Delta G^\circ$) The relationship between standard Gibbs free energy change ($\Delta G^\circ$), standard enthalpy change ($\Delta H^\circ$), and standard entropy change ($\Delta S^\circ$) is defined by the Gibbs equation: \[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\] Substitute the given values into the equation: * $\Delta H^\circ = 18000\text{ cal}$ * $T = 300\text{ K}$ * $\Delta S^\circ = 30\text{ cal K}^{-1}$ \[\Delta G^\circ = 18000\text{ cal} - (300\text{ K} \times 30\text{ cal K}^{-1})\] \[\Delta G^\circ = 18000\text{ cal} - 9000\text{ cal}\] \[\Delta G^\circ = 9000\text{ cal}\] Step 3 - Calculate the standard equilibrium constant ($K_p^\circ$) The standard Gibbs free energy change is related to the thermodynamic equilibrium constant ($K_p^\circ$) by the following equation: \[\Delta G^\circ = -RT \ln K_p^\circ\] Rearranging the expression to solve for $\ln K_p^\circ$: \[\ln K_p^\circ = -\frac{\Delta G^\circ}{RT}\] Substituting the values, where the universal gas constant $R \approx 2\text{ cal K}^{-1}\text{ mol}^{-1}$ (more precisely $1.987\text{ cal K}^{-1}\text{ mol}^{-1}$): \[\ln K_p^\circ = -\frac{9000\text{ cal}}{2\text{ cal K}^{-1}\text{ mol}^{-1} \times 300\text{ K}}\] \[\ln K_p^\circ = -\frac{9000}{600}\] \[\ln K_p^\circ = -15\] Now, take the exponential on both sides to solve for $K_p^\circ$: \[K_p^\circ = \boxed{e^{-15}}\] Step 4 - Evaluate the options * **Option (A) $e^{15}$**: Incorrect. This would be the value if the negative sign in the Gibbs-equilibrium constant relationship was omitted, resulting in $\ln K_p^\circ = +15$. * **Option (B) $e^{-15}$**: Correct. Applying the correct thermodynamic constants and algebraic signs yields $e^{-15}$. * **Option (C) $e^{-18}$**: Incorrect. This corresponds to an error in calculating Gibbs free energy where the entropy term is neglected, or from a calculation error. * **Option (D) $e^{-12}$**: Incorrect. This corresponds to an arithmetic error during the subtraction or division process.