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17 January, 01:05

The second-order rate constant for the reaction A + 2 B - >C + D is 0.34 dm' mo1·• s'. / /'hat is the concentration of C after (i) 20s, (ii) 15 mi n when the reactants are mixed with initial concentrations of [A] = 0.027 mol dm • 1 and [BJ = 0.130mol dm""?

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  1. 17 January, 01:23
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    i) After 20 s the concentration of C is 0.024 M

    ii) after 15 min the concentration of C is 0.027 M

    Explanation:

    Let's assume that the reaction is first order for each reactive, making a second order global reaction. In that case:

    v = k[A][B] = 0.34 M⁻¹ s⁻¹ * 0.027 M * 0.130 M = 1.2 x 10⁻³ M / s

    First, let's see how much time it takes for the reactants to disappear:

    The rate of disappearance of A will be:

    -Δ[A] / Δt = v

    where:

    Δ[A] = final - initial concentration of A

    Δ[t] = elapsed time

    Then:

    Δ[t] = - Δ[A] / v

    Δ[t] = - (-0.027 M) / 1.2 x 10⁻³ M / s = 22.5 s

    The rate of disappearance of B will be:

    -1/2Δ[B] / Δt = v

    Δt = - 1/2 * (-0.130 M) / 1.2 x 10⁻³ M / s

    Δt = 54.2 s

    Then, the reactant A will completely disappear in 22.5 s and it will limit the reaction.

    The rate of production of C will be:

    Δ[C] / Δt = v

    where

    Δ[C] = final concentration of C - initial concentration of C

    Δt = final time - initial time

    v = rate of the reaction

    Then:

    Δ[C] = v * Δt

    Since the initial concentration of C is 0 and the initial time is 0, then:

    [C] = v * t

    i) t = 20 s

    [C] = 1.2 x 10⁻³ M / s * 20 s = 0.024 M

    ii) t = 15 min = 15 min * (60 s / 1 min) = 900 s

    The reaction does not occur beyond the 22.5 s which is the time in which A disappears. Then, the concentration of C at 15 min will be the same as the concentration at 22.5 s:

    [C] = 1.2 x 10⁻³ M / s * 22.5 s = 0.027 M
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