Import Question JSON

$\text{HINT: At equilibrium } \Delta\text{G} = 0$ $\text{Step 1: Understand the condition for equilibrium}$ $\text{At equilibrium, the Gibbs free energy change (}\Delta\text{G) for the reaction is zero. This is a fundamental principle in thermodynamics that indicates the system has reached a state where there is no net tendency for further change.}$ $\text{Step 2: Apply the Gibbs free energy equation}$ $\text{The Gibbs free energy change for a reaction is related to the enthalpy change (}\Delta\text{H) and entropy change (}\Delta\text{S) by the equation:}$ $\Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S}$ $\text{Where:}$ $\Delta\text{G} = \text{Gibbs free energy change}$ $\Delta\text{H} = \text{enthalpy change}$ $\Delta\text{S} = \text{entropy change}$ $\text{T} = \text{temperature in Kelvin}$ $\text{Step 3: Substitute the known values and solve for the equilibrium temperature}$ $\text{Given:}$ $\Delta\text{H} = 30 \text{ kJ mol}^{-1} = 30,000 \text{ J mol}^{-1}$ $\Delta\text{S} = 105 \text{ J K}^{-1} \text{ mol}^{-1}$ $\text{At equilibrium, } \Delta\text{G} = 0$ $\text{Therefore:}$ $0 = \Delta\text{H} - \text{T}\Delta\text{S}$ $\text{Rearranging to solve for T:}$ $\text{T}\Delta\text{S} = \Delta\text{H}$ $\text{T} = \frac{\Delta\text{H}}{\Delta\text{S}}$ $\text{Substituting the values:}$ $\text{T} = \frac{30,000 \text{ J mol}^{-1}}{105 \text{ J K}^{-1} \text{ mol}^{-1}}$ $\text{T} = 285.7 \text{ K}$ $\text{This is the temperature at which the reaction will be in equilibrium.}$ $\text{Physical interpretation: At this temperature (285.7 K), the reaction is at equilibrium, meaning there is no net conversion of reactants to products or vice versa. Above this temperature, the reaction will favor the formation of products (BrCl) because the T}\Delta\text{S term becomes larger, making }\Delta\text{G negative. Below this temperature, the reaction will favor the reactants (Br}_2 \text{ and Cl}_2\text{) because the }\Delta\text{H term dominates, making }\Delta\text{G positive.}$