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In a reaction, A+BProduct, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law - Chemistry (Theory)

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प्रश्न

In a reaction, \[\ce{A + B -> Product}\], rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as ______.

विकल्प

  • Rate = k[A][B]2

  • Rate = k[A]2[B]2

  • Rate = k[A][B]

  • Rate = k[A]2[B]

MCQ
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उत्तर

In a reaction, \[\ce{A + B -> Product}\], rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as Rate = k[A]2[B].

Explanation:

The given reaction is 

\[\ce{A + B -> Product}\]

When [B] is doubled, rate is doubled

⇒ Order w.r.t B = 1

When [A] and [B] are both doubled, rate increases 8 times.

We already know doubling [B] alone doubles the rate, so the additional increase must be due to [A]:

Total rate increase = 8

From B alone = 2

From A alone = `8/2` = 4

So, 2n = 4

n = `4/2`

n = 2

⇒ Order w.r.t. A = 2

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अध्याय 4: Chemical Kinetics - OBJECTIVE (MULTIPLE CHOICE) TYPE QUESTIONS [पृष्ठ २६८]

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नूतन Chemistry Part 1 and 2 [English] Class 12 ISC
अध्याय 4 Chemical Kinetics
OBJECTIVE (MULTIPLE CHOICE) TYPE QUESTIONS | Q 34. | पृष्ठ २६८
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