Question

When heated, iodine vapor dissociates according to this equation:

a) I2(g) ⇌ I(g)
b) I(g) ⇌ I2(g)
c) 2I(g) ⇌ I2(g)
d) I(g) + I2(g) ⇌ I3(g)

Answer 1

The dissociation of iodine vapor when heated is accurately represented by the equation 2I(g) ⇌ I2(g), which is option (c), illustrating a dynamic equilibrium between iodine atoms and molecules.

Step-by-step explanation:

When heated, iodine vapor dissociates predominantly through the equation 2I(g) ⇌ I2(g). This dissociation is reversible and is an example of a dynamic equilibrium where iodine molecules (I2) in the gas phase break apart into individual iodine atoms (I) and also recombine.

At higher temperatures, more iodine molecules dissociate into atoms due to the increased kinetic energy, while at lower temperatures, the atoms recombine into molecules more frequently.

The correct option that describes the dissociation of iodine vapor when heated is therefore option (c): 2I(g) ⇌ I2(g). This represents a classic case of dynamic equilibrium where the rate of the forward reaction, which is the dissociation of iodine molecule into iodine atoms, is equal to the rate of the reverse reaction, the recombination of iodine atoms into iodine molecules.