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The rate of the reaction NO 2 (g)+ CO(g) --> NO(g) + CO(g) depends only on the c

ID: 677016 • Letter: T

Question

The rate of the reaction               NO2(g)+ CO(g) --> NO(g) + CO(g) depends only on the concentration of the nitrogen dioxidebelow 225 degrees C. At a temperature below 225 degrees C, the following data werecollected. Time (s)      [NO2](mol/L)    0                  .500 1.2 x103            .444 3.0 x103            .381 4.5 x103            .340 9.0 x103            .250 1.8 x104            .174 Determine the rate law, the integrated law, and the value ofthe rate constant. Calculate [NO2] at 2.70 x 104 seconds after the start of thereaction. The rate of the reaction               NO2(g)+ CO(g) --> NO(g) + CO(g) depends only on the concentration of the nitrogen dioxidebelow 225 degrees C. At a temperature below 225 degrees C, the following data werecollected. Time (s)      [NO2](mol/L)    0                  .500 1.2 x103            .444 3.0 x103            .381 4.5 x103            .340 9.0 x103            .250 1.8 x104            .174 Determine the rate law, the integrated law, and the value ofthe rate constant. Calculate [NO2] at 2.70 x 104 seconds after the start of thereaction.

Explanation / Answer

   We Know that :           Thegiven equation is :                NO2(g)+ CO(g) --> NO(g) + CO2(g)              By plotting the graph for ln [ NO2 ] &Time and 1/ [ NO2 ] vstimte we get straight line                  Which indicates that :                  Rate = - [ NO2 ] / t   = K [ NO2 ]2       Rate  constant of theReaction = 1 / NO2   fromthe final time to the initial time                                         K   = 5.7471 - 2 / 1800 - 0                                               =   0.00208173L / mol-s          IntegratedRate law for the second order reaction is :            1/ [ NO2 ] = Kt + 1 / [NO2 ]0         Rate of theReaction for IInd Order is :             1 / [ NO2] = 2.08 x 10-4 L/ mol-s x 27000 s + 1 / 0.500 M                                 = 7.616                  [ NO2 ]   = 0.13130 M         Rate of theReaction for IInd Order is :             1 / [ NO2] = 2.08 x 10-4 L/ mol-s x 27000 s + 1 / 0.500 M                                 = 7.616                  [ NO2 ]   = 0.13130 M           
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