Question

Consider the balanced chemical reaction,

2HI(g)-->H2(g) + I2(g)

If the rate of change of HI (∆[HI]/∆t) is −1.9 x 10^−4 M/s at a particular time, what is rate of the reaction at the same time?

Answer 1

The rate of the reaction, given the rate of change of HI, is -9.5 x 10⁻⁵ M/s, which is determined by dividing the rate of HI consumption by 2 due to the stoichiometry of the balanced chemical equation.

Step-by-step explanation:

The question involves the balanced chemical reaction 2HI(g) → H₂(g) + I₂(g) and asks for the rate of the reaction given the rate of change of HI (∆[HI]/∆t).

Given the reaction rate for HI is -1.9 x 10⁻⁴ M/s, the rate of disappearance of HI is twice that of the formation rates of H₂ and I₂ because of the stoichiometry of the balanced chemical equation. Both H₂ and I₂ are produced at half the rate that HI is consumed. Therefore, the rate of the reaction, which is the change in concentration of H₂ or I₂ over time, will be -1.9 x 10⁻⁴ M/s divided by 2, which is -9.5 x 10⁻⁵ M/s. This is because for every 2 moles of HI that react, 1 mole of H₂ and 1 mole of I₂ are produced.