Question

Potassium chlorate, kclo3, decomposes according to the reaction: 2kclo3 → 2kcl 3o2 if the rate of decomposition of kclo3 at a certain time is determined to be 2.4 x 10-2 mol s-1, what is the rate of formation of kcl at the same time

Answer 1

The rate of formation of KCl is the same as the rate of decomposition of KClO₃, which is 2.4 × 10⁻² mol s⁻¹.

Step-by-step explanation:

The rate of decomposition of Potassium chlorate, or KClO₃, is given as 2.4 × 10⁻² mol s⁻¹. Based on the balanced chemical equation provided, 2KClO₃ → 2KCl + 3O₂, there is a 1:1 molar ratio between KClO₃ and KCl. Therefore, if KClO₃ is decomposing at a rate of 2.4 × 10⁻² mol s⁻¹, then KCl is being formed at the same rate, which is also 2.4 × 10⁻² mol s⁻¹, since both KClO₃ and KCl have a coefficient of 2 in the balanced equation, indicating equal amounts change per unit time.

Answer 2

`2.4 * 10^(-2) \; \text{mol} \cdot \text{s}^(-1)`

Explanation

Referring to coefficients of
`\text{KClO}_3` and
`\text{KCl}` the equation, the decomposition of every
`2` moles of potassium chlorate would yield
`2` moles of potassium chloride. At any time in the reaction process the rate of production of
`\text{KCl}` shall equals to the rate at which
`\text{KClO}_3` is being consumed.

The question states that the reaction consumes
`2.4 * 10^(2) \; \text{mol}` of
`\text{KClO}_3` every single second on average. One would thus expect it to produce the same amount of
`\text{KCl}` in the same period of time.
`\text{KCl}` is thus produced at a rate of
`2.4 * 10^(2) \; \text{mol} \cdot \text{s}^(-1)`