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

If the initial concentration of X is 1.0 M, what is the concentration of X after 195 s?

a.

0.19 M

b.

0.98 M

c.

0.016 M

d.

60 M

e.

0.59 M

### 5 Answers

- wanszetoLv 71 month ago
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)

====

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)

- hcbiochemLv 71 month ago
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

- JimLv 71 month ago
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.

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- JakeLv 41 month ago
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