For the reaction 3A(g) +2B(g) --> 2C(g) + 2D(g) the following data was collected
ID: 683265 • Letter: F
Question
For the reaction 3A(g) +2B(g) --> 2C(g) + 2D(g) the following data was collected at constanttemperature. Determine the correct rate law for thisreaction. and Why. Trial Initial[A] Initial [B] InitialRate (mol/L) (mol/L) (mol/Lxmin) 1 0.200 0.100 6.00x 10-2 2 0.100 0.100 1.50x 10-2 3 0.200 0.200 1.20x 10-1 4 0.300 0.200 2.70x 10-1 a) rate = k[A][B] b) rate = k[A][B]2 c) rate = k[A]3[B]2 d) rate = k[A]1.5[B] e) rate = k[A]2[B] For the reaction 3A(g) +2B(g) --> 2C(g) + 2D(g) the following data was collected at constanttemperature. Determine the correct rate law for thisreaction. and Why. Trial Initial[A] Initial [B] InitialRate (mol/L) (mol/L) (mol/Lxmin) 1 0.200 0.100 6.00x 10-2 2 0.100 0.100 1.50x 10-2 3 0.200 0.200 1.20x 10-1 4 0.300 0.200 2.70x 10-1 a) rate = k[A][B] b) rate = k[A][B]2 c) rate = k[A]3[B]2 d) rate = k[A]1.5[B] e) rate = k[A]2[B] d) rate = k[A]1.5[B] e) rate = k[A]2[B]Explanation / Answer
For A, when the concentration of A is doubled while B remainsconstant, the rate is quadrupled. Therefore, the order of A is 2.(compare rows 1 & 2) For B, when the concentration of B is doubled while keeping Aconstant, the rate also doubles. So the order of B is 1. (comparerows 1 & 3) The answer is E.
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