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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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