Suppose the formation of iodine proceeds by the following mechanism:

step elementary reaction rate constant
1 H2 (g) + ICl (g) → HI (g) + HCl (g) k1
2 HI (g) + ICl (g) → I2 (g) + HCl (g) k2

Suppose also ki « k2. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction. Write the experimentally- observable rate law for the overall chemical reaction.

Overall reaction

H2(g) + 2ICI(g) > I2(g) +2HCl(g)

Overall Rate = k1[H2] [ICl]

Explanation:

Overall reaction

H2(g) + 2ICI(g) > I2(g) +2HCl(g)

The overall reaction is the sum of the two two reactions shown in the question. After the two reactions are summed up properly, this overall reaction equation his obtained.

Since K1<<K2 it means that step 1 is slower than step 2. Recall that the rate if reaction depends on the slowest step of the reaction. Hence

Overall Rate = k1[H2] [ICl]

it gains energy in a quantized amount

Explanation:

when we describe the energy of a particle as a quantized ,we mean that only certain values of energy are allowedit can only gain the exact amount of energy needed to reach one of the higher energy levels

hope this helps :)