Relate the rate of: 1 disappearance of iodine vapour to the rate of formation of hydrogen iodide for the following reaction: H2(g) +I2(g) _ 2HI(g)?

The rate of disappearance of iodine vapor is directly related to the rate of formation of hydrogen iodide in the given reaction. This is because the disappearance of iodine vapor is a result of its reaction with hydrogen gas to form hydrogen iodide. As hydrogen iodide is formed, iodine vapor is consumed, leading to a decrease in its concentration. Therefore, the rate at which iodine vapor disappears is directly proportional to the rate at which hydrogen iodide is formed in the reaction.

The rate of disappearance of iodine vapour is equal to the rate of formation of hydrogen iodide.

Rate of disappearance of iodine vapour = - d[I2]/dt, where [I2] is the concentration of iodine vapour at time t.

Rate of formation of hydrogen iodide = d[HI]/dt, where [HI] is the concentration of hydrogen iodide at time t.

According to the stoichiometry of the reaction, for every mole of I2 that disappears, two moles of HI are formed. Therefore,

- d[I2]/dt = 2 - d[HI]/dt

Rearranging the equation:

d[HI]/dt = -1/2 - d[I2]/dt

Therefore, the rate of formation of hydrogen iodide is equal to half the negative of the rate of disappearance of iodine vapour.