What is Δ_rH° for a reaction that has K_p = 1.456 at 273 K and K_p = 14.2 at 298 K? A)8.42 × 10^-5 kJ/mol B)61.6 kJ/mol C)-61.6 kJ/mol D)462 kJ/mol E)-462 kJ/mol

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The change in standard Gibbs free energy (ΔG°) can be related to the equilibrium constant (K) and temperature (T) by the equation ΔG° = -RTlnK, where R is the gas constant (8.314 J/mol·K). The change in ΔG° with temperature gives us the standard enthalpy change (ΔH°) for the reaction, according to the van't Hoff equation: $\frac{d(lnK)}{dT} = \frac{\Delta H°}{RT^2}$. Given the values of K at two different temperatures, we can solve for ΔH°. The negative sign in the choice indicates a direction opposite to what the calculation would yield for an exothermic reaction, and the magnitude of the value in choice B is reasonable for chemical reactions, making it the correct choice.

What Is ΔrH° for a Reaction That Has Kp =

What Is Δ_rH° for a Reaction That Has K_p =