Answer:
the rate constant (k) for the decomposition system is approximately 0.173 min-1
Explanation:
For a reaction that follows first-order kinetics, the rate equation is given by:
Ratek [Reactant]
The integrated rate law for a first-order reaction is:
In Reactant Reactants -kt
Where:
*k is the rate constant,
[Reactant], is the concentration of the reactant at time t,
. [Reactant] is the initial concentration of the reactant.
In this case, the concentration is reduced to of its initial value in 12 minutes. Let [Reactant], be the final concentration, [Reactant] be the initial concentration, and t be the time.
The given information can be expressed as:
In Reactant Reactant =-kt
Given that Reactant Reactant andt = 12 minutes:
-k-12
Now, solve for k:
-2.079-12k
-2079 12
k≈ 0.173 min-1
Therefore, the rate constant (k) for the decomposition system is approximately 0.173 min-1
The rate constant k for a first-order reaction where concentration is reduced to 1/8 in 12 minutes is calculated using the first-order kinetics equation. Convert the time to seconds, then calculate k using ln(8) = k * (12 * 60).
To determine the rate constant of the decomposition system where the concentration is reduced to 1/8 of its initial value in 12 minutes, we utilize the equation for first-order kinetics: ln([A]0/[A]) = kt. Let [A]0 be the initial concentration and [A] the final concentration after time t. Given that [A] is 1/8 of [A]0, we have ln([A]0/(1/8)[A]0) = ln(8) = kt.
Using the provided time t of 12 minutes (which must be converted to seconds by multiplying it by 60 to match the units of rate constant k, which are typically in seconds), we can compute the rate constant k: ln(8) = k * (12 * 60). By calculating, we find that the rate constant k is equal to: k = ln(8) / (12 * 60), Now, k can be solved and will be in the units of s-1.
