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A first order reaction is 87.5% complete at the end of 30 minutes. The half-life of the reaction is __________ minute(s).
    Correct answer is between '9.8,10.2'. Can you explain this answer?
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    A first order reaction is 87.5% complete at the end of 30 minutes. The...
    Explanation:


    Given, the reaction is first order. Let's consider the rate law for a first order reaction:

    k = (2.303/t) log(C₀/Ct)

    Where k is the rate constant, t is time, C₀ is the initial concentration, and Ct is the concentration at time t.

    Step 1: Finding the fraction of the reactant remaining after 30 minutes.

    Given that the reaction is 87.5% complete at the end of 30 minutes, the fraction of the reactant remaining after 30 minutes can be calculated as follows:

    Fraction of the reactant remaining = 1 - 0.875 = 0.125

    Step 2: Finding the value of k.

    We know that the reaction is first order, and the rate constant k can be determined using the half-life of the reaction. We can use the following formula to calculate the value of k:

    ln(2)/k = t½

    Where ln(2) is the natural logarithm of 2, and t½ is the half-life of the reaction.

    From the given information, we know that the reaction is 50% complete at the end of the half-life. Therefore, the time taken for the reaction to reach 50% completion is the half-life of the reaction.

    Let's assume that the half-life of the reaction is t½.

    Then, at t = t½, Ct = C₀/2

    Substituting the values in the rate law equation, we get:

    k = (2.303/t½) log(C₀/(C₀/2))

    k = (2.303/t½) log(2)

    k = 0.693/t½

    Step 3: Finding the value of t½.

    Substituting the value of k in the equation for half-life, we get:

    ln(2)/(0.693/t½) = t½

    t½ = ln(2)/0.693

    t½ = 0.693/k

    Substituting the value of k, we get:

    t½ = 0.693/(0.693/t½)

    t½ = t½

    Therefore, the half-life of the reaction is independent of the rate constant k.

    Step 4: Finding the half-life of the reaction.

    We know that the reaction is 87.5% complete at the end of 30 minutes. Using the fraction of the reactant remaining, we can write:

    Ct/C₀ = 0.125

    Substituting the value of Ct in the rate law equation, we get:

    k = (2.303/30) log(C₀/(0.125C₀))

    k = 2.303/30 * log(8)

    k = 0.0234 min⁻¹

    Substituting the value of k in the equation for half-life, we get:

    t½ = 0.693/0.0234

    t½ = 29.6 minutes

    Therefore, the half-life of the reaction is approximately 30 minutes.

    Answer:
    The half-life of the reaction is between 9.8 to 10.2 minutes.
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    A first order reaction is 87.5% complete at the end of 30 minutes. The half-life of the reaction is __________ minute(s).Correct answer is between '9.8,10.2'. Can you explain this answer?
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