Final answer:
The equilibrium constant for the reaction H2 + I2 ⇔ 2 HI can be calculated using an ICE table and the equilibrium concentrations of H2, I2, and HI. Additional information about the initial moles of reactants is required to solve for K.
Step-by-step explanation:
The question concerns the calculation of the equilibrium constant (K) for the chemical reaction between hydrogen (H2) and iodine (I2) to form hydrogen iodide (HI) at a temperature of 765 K in a 5-litre volume container. Given that we have 0.4 moles of HI at equilibrium, we can set up an ICE table (Initial, Change, Equilibrium) to determine the moles of H2 and I2 at equilibrium and then calculate the equilibrium concentrations. These concentrations can then be used to calculate the equilibrium constant using the formula K = [HI]2/([H2]*[I2]).
Let's assume x moles of H2 and I2 react to form HI. At equilibrium, we will have (0.4 moles - 2x) of HI, x moles of H2, and x moles of I2. The equilibrium concentrations will be those moles divided by the volume of the container. Plugging the concentrations into the equilibrium constant formula and solving for K will lead us to the correct answer. Note that to determine the exact value of K, we'd need additional information such as the initial moles or concentrations of H2 and I2, which are not provided in the question.