The initial rates method was used to study the reactionbelow. A + 3B ® 2C [A](mo

Question

The initial rates method was used to study the reactionbelow.

                             A + 3B ® 2C

      [A](mol/L)                 [B](mol/L)                 -D[A]/Dt (mol/L·s)

     0.210                          0.150                          3.41 × 10-3

     0.210                          0.300                          1.36 × 10-2

     0.420                          0.300                          2.73 × 10-2

Determine the rate expression and calculate the rate constantfor the reaction.

*can i please get another opinion!!! pleaseeee!!!

Explanation / Answer

compare run1 and run 2 rate2/rate1 = (1.36e-2/3.41e-3) = 3.988 ~ 4 So doubling the B cocentration quadruples the rate  => rate is proporitonal to [B]2 compare run3 and run2 rate increases by a factor of 2 when you double [A]  =>   rate is proportional to [A] overall rate = k*[A][B]2 k = rate / [A][B]2 = [2.73e-2 M/s]/[0.420 M* 0.300M2] = 0.722 M-2 s-1

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