(a) Is the dissociation of fluorine molecules into atomic fluorine, \(\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{~F}(g)\) an exothermic or endothermic process? (b) If the temperature is raised by \(100 \mathrm{~K}\), does the equilibrium constant for this reaction increase or decrease? (c) If the temperature is raised by \(100 \mathrm{~K},\) does the forward rate constant \(k_{f}\) increase by a larger or smaller amount than the reverse rate constant \(k_{r} ?\)

Since breaking chemical bonds requires energy, dissociation of fluorine molecules into atomic fluorine requires an input of energy. Therefore, the given reaction is an endothermic process. Answer (a): The dissociation of fluorine molecules into atomic fluorine is an endothermic process.

According to Le Châtelier's principle, if we increase the temperature of an endothermic reaction, the equilibrium will shift towards the endothermic side (the right side) in order to absorb the added energy. As the reaction shifts to the right, the concentration of products (atomic fluorine) will increase, leading to an increase in the equilibrium constant. Answer (b): If the temperature is raised by 100 K, the equilibrium constant for this reaction will increase.

According to the Arrhenius equation, the rate constants of both forward and reverse reactions will increase with an increase in temperature: \(k = Ae^{-\frac{Ea}{RT}}\) Where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. As the temperature increases, the exponent term becomes smaller, and k increases. However, the increase in the forward rate constant (kf) will be greater than that of the reverse rate constant (kr) since the forward reaction (an endothermic process) better absorbs the added thermal energy than the reverse reaction (an exothermic process). Answer (c): If the temperature is raised by 100 K, the forward rate constant (kf) increases by a larger amount than the reverse rate constant (kr).