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Chemical Kinetics question

2010 · Q107
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Chemical Kinetics question

2010 · Q107

NEETChemistryChemical KineticsMCQ+4 / −1
During the kinetic study of the reaction, 2A + B →\to→ C + D, following results were obtained
Run [A]/mol L−-−1 [B]/mol L−-−1 Initial rate of formation
of D/mol L−-−1 min−-−1
I. 0.1 0.1 6.0×\times×10−-−3
II. 0.3 0.2 7.2×\times×10−-−2
III. 0.3 0.4 2.88×\times×10−-−1
IV. 0.4 0.1 2.40×\times×10−-−2

Based on the above data which one of the following is correct?
  1. A
    Rate = k[A]2[B]
  2. B
    Rate = k[A][B]
  3. C
    Rate = k[A]2[B]2
  4. D
    Rate = k[A][B]2
View written solutionFree

Correct answer: D

Rate = k[A]x [B]y

For the given situations

(I) rate = k(0.1)x (0.1)y = 6.0×\times×10−-−3

(II) rate = k(0.2)x (0.3)y = 7.2×\times×10−-−2

(III) rate = k(0.3)x (0.4)y = 2.88×\times×10−-−1

(IV) rate = k(0.4)x (0.1)y = 2.40×\times×10−-−2

Dividing eq. (I) by eq. (IV) we get

(0.10.4)x(0.10.1)y=6.0×10−32.4×10−2{\left( {{{0.1} \over {0.4}}} \right)^x}{\left( {{{0.1} \over {0.1}}} \right)^y} = {{6.0 \times {{10}^{ - 3}}} \over {2.4 \times {{10}^{ - 2}}}}(0.40.1​)x(0.10.1​)y=2.4×10−26.0×10−3​

⇒\Rightarrow⇒ (14)x=(14)1{\left( {{1 \over 4}} \right)^x} = {\left( {{1 \over 4}} \right)^1}(41​)x=(41​)1

⇒\Rightarrow⇒ x = 1

On dividing eq. (II) by eq. (III) we get

(0.30.3)x(0.20.4)y=7.2×10−22.88×10−1{\left( {{{0.3} \over {0.3}}} \right)^x}{\left( {{{0.2} \over {0.4}}} \right)^y} = {{7.2 \times {{10}^{ - 2}}} \over {2.88 \times {{10}^{ - 1}}}}(0.30.3​)x(0.40.2​)y=2.88×10−17.2×10−2​

⇒\Rightarrow⇒ (12)y=14{\left( {{1 \over 2}} \right)^y} = {1 \over 4}(21​)y=41​

⇒\Rightarrow⇒ y = 2

∴\therefore∴ Rate = k[A][B]2

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