A chemical reaction that is first order in X is observed to have a rate constant of 2.1 × 10 –2s –1. ?

Method 1:

The integrated rate law for first order reaction:
ln[X] = -kt + ln[X]ₒ
ln[X] = -(2.1 × 10⁻²) × 195 + ln(1.0)
ln[X] = -4.095
Concentration of X after 195 s, [X] = e⁻⁴·⁰⁹⁵ M = 0.017 M

The answer: c. 0.016 M (the closest answer)

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Method 2:

Half-live, t½ = ln(2)/k = ln(2)/ (2.1 × 10⁻²) = 33 s
No. of half-lives = (195 s) / (33 s) = 5.91
Concentration of X after 195 s, [X] = (1.0 M) × (1/2)⁵·⁹¹ = 0.017 M

The answer: c. 0.016 M (the closest answer)

Use the integrated rate law for a first order reaction:

ln [X = -kt + ln [X]o
ln [X] = -2.1X10^-2 (195)s + ln 1.0
ln [X] = -4.095
[X] = 0.017 M

rate constant of  2.1 × 10^(–2)/s
195 s (2.1 × 10^(–2)/s) = 4.095 conc change
Reaction rate is calculated using the formula rate = Δ[C]/Δt, where Δ[C] is the change in product concentration during time period Δt.

"2.1 × 10 –2s –1" ?? what does this mean?

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