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Chemical Kinetics and Nuclear Chemistry question

2024 · 5 Apr · Shift 1 · Q30
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Chemical Kinetics and Nuclear Chemistry question

2024 · 5 Apr · Shift 1 · Q30

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1

During Kinetic study of reaction 2A+B→C+D\mathrm{2 A+B \rightarrow C+D}2A+B→C+D, the following results were obtained :

A[M]\mathrm{A [M]}A[M] B[M]\mathrm{B [M]}B[M] initial rate of formation of D\mathrm{D}D
I 0.1 0.1 6.0×10−36.0\times10^{-3}6.0×10−3
II 0.3 0.2 7.2×10−27.2\times10^{-2}7.2×10−2
III 0.3 0.4 2.88×10−12.88\times10^{-1}2.88×10−1
IV 0.4 0.1 2.40×10−22.40\times10^{-2}2.40×10−2

Based on above data, overall order of the reaction is ‾\underline{\hspace{2cm}}​.

Numerical answer
View written solutionFree

Correct answer: 3

  1. Assume the rate law

    For the reaction 2A+B→C+D2A + B \rightarrow C + D2A+B→C+D let the rate of formation of DDD be r=k[A]m[B]nr = k[A]^m[B]^nr=k[A]m[B]n where mmm and nnn are the orders with respect to AAA and BBB.

  2. Find order with respect to AAA

    Compare experiments I and IV:

    • In I: [A]=0.1[A]=0.1[A]=0.1, [B]=0.1[B]=0.1[B]=0.1, rate =6.0×10−3=6.0\times 10^{-3}=6.0×10−3
    • In IV: [A]=0.4[A]=0.4[A]=0.4, [B]=0.1[B]=0.1[B]=0.1, rate =2.40×10−2=2.40\times 10^{-2}=2.40×10−2

    Here, [B][B][B] is constant.

    So, r4r1=([A]4[A]1)m\frac{r_4}{r_1} = \left(\frac{[A]_4}{[A]_1}\right)^mr1​r4​​=([A]1​[A]4​​)m

    2.40×10−26.0×10−3=(0.40.1)m\frac{2.40\times 10^{-2}}{6.0\times 10^{-3}} = \left(\frac{0.4}{0.1}\right)^m6.0×10−32.40×10−2​=(0.10.4​)m

    4=4m4 = 4^m4=4m

    Hence, m=1m=1m=1

  3. Find order with respect to BBB

    Compare experiments II and III:

    • In II: [A]=0.3[A]=0.3[A]=0.3, [B]=0.2[B]=0.2[B]=0.2, rate =7.2×10−2=7.2\times 10^{-2}=7.2×10−2
    • In III: [A]=0.3[A]=0.3[A]=0.3, [B]=0.4[B]=0.4[B]=0.4, rate =2.88×10−1=2.88\times 10^{-1}=2.88×10−1

    Here, [A][A][A] is constant.

    So, r3r2=([B]3[B]2)n\frac{r_3}{r_2} = \left(\frac{[B]_3}{[B]_2}\right)^nr2​r3​​=([B]2​[B]3​​)n

    2.88×10−17.2×10−2=(0.40.2)n\frac{2.88\times 10^{-1}}{7.2\times 10^{-2}} = \left(\frac{0.4}{0.2}\right)^n7.2×10−22.88×10−1​=(0.20.4​)n

    4=2n4 = 2^n4=2n

    Hence, n=2n=2n=2

  4. Overall order

    Overall order=m+n=1+2=3\text{Overall order} = m+n = 1+2 = 3Overall order=m+n=1+2=3

  5. Comparison with stored answer

    Derived answer = 333

    Stored correct answer = 333

    So they agree.

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