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

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  • 4 weeks 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)

  • 4 weeks 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

  • 4 weeks ago

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

  • Jim
    Lv 7
    4 weeks 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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  • Jake
    Lv 4
    4 weeks ago

    The Virgin Mary is Our Lady of Perpetual Help.  She can help one with any dilemma in life.  It is good to honor her every day by reciting the rosary with care and sincerity.

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