A graph of ΔG° (J mol⁻¹) versus ln K for a reaction at constant temperature passes through the following two points:
(ln K = 0.00, ΔG° = 0 J mol⁻¹) and (ln K = 1.80, ΔG° = –4.5 × 10³ J mol⁻¹).
(1) Determine the temperature, in kelvins, at which the reaction was studied.
(2) Using the temperature found in part (a), predict the value of ΔG° (in kJ mol⁻¹) when K = 50.