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Question: For the reaction $3A(g) \xrightarrow{k} B(g) + C(g)$ k is $10^{-4}L/mol.min$. If $[A] = 0.5M$ then t...

For the reaction 3A(g)→kB(g)+C(g)3A(g) \xrightarrow{k} B(g) + C(g) k is 10−4L/mol.min10^{-4}L/mol.min. If [A]=0.5M[A] = 0.5M then the value of −d[A]dt(in ms−1)-\frac{d[A]}{dt} (in \ ms^{-1}) is:

A

7.5 x 10−510^{-5}

B

3 x 10−410^{-4}

C

2.5 x 10−510^{-5}

D

1.25 x 10−610^{-6}

Answer

1.25 x 10−610^{-6} M/s

Explanation

Solution

Solution:

  1. Determine the rate law:
    The given rate constant k=10−4 L/(mol⋅min)k = 10^{-4}\,L/(mol\cdot min) has units of L mol−1 min−1L\,mol^{-1}\,min^{-1}. For a rate law of the form

    Rate=k[A]n,\text{Rate} = k[A]^n,

    we require the overall order n=2n=2 (since for a second-order reaction, the units of kk are L mol−1 min−1L\,mol^{-1}\,min^{-1}). Thus, the reaction is second order in [A][A] even though the stoichiometric coefficient is 3.

  2. Relate the reaction rate to the disappearance of AA:
    For the elementary reaction

    3A⟶B+C,3A \longrightarrow B + C,

    the rate of the reaction is given by

    Rate=−13d[A]dt.\text{Rate} = -\frac{1}{3} \frac{d[A]}{dt}.

    Equate this with the rate law:

    k[A]2=−13d[A]dt⟹−d[A]dt=3k[A]2.k[A]^2 = -\frac{1}{3}\frac{d[A]}{dt} \quad \Longrightarrow \quad -\frac{d[A]}{dt} = 3k[A]^2.

  3. Substitute the given values:
    With [A]=0.5 M[A] = 0.5\,M:

    −d[A]dt=3×10−4×(0.5)2=3×10−4×0.25=7.5×10−5 M/min.-\frac{d[A]}{dt} = 3\times10^{-4}\times (0.5)^2 = 3\times10^{-4}\times 0.25 = 7.5\times10^{-5}\, \text{M/min}.

  4. Convert the rate to Moles per Liter per Second (M/s):
    Since 1 min=60 s1\,min = 60\,s,

    7.5×10−5 M/min=7.5×10−560 M/s=1.25×10−6 M/s.7.5\times10^{-5}\, \text{M/min} = \frac{7.5\times10^{-5}}{60}\, \text{M/s} = 1.25\times10^{-6}\, \text{M/s}.