Two isomers (A and B) of a given compound dimerize as follows. 2 A A2 2 B B2 Bot

Question

Two isomers (A and B) of a given compound dimerize as follows. 2 A A2 2 B B2 Both processes are known to be second order in reactants, andk1 is known to be 0.221 L/mol s at 28 degree C. In a particular experiment A and B were placedin separate containers at 28 degree C, where[A]0 = 1.39M and [B]0 = 2.81 M. It was found thatafter each reaction has progressed for 2.95 mins [A] = 2.95[B]. In this case the rate laws are defined asfollows. Rate = = k1[A]2 Rate = = k2[B]2 (a) Calculate the concentration of A2 after 2.95 min. M (b) Calculate the value of k2. L/mol s (c) Calculate the half-life for the experiment involving A.

Explanation / Answer

We Know that according to second order Rate - Law:     1 / [ A ] = Kt + 1 / [ A ]0      1 / [ A2]   = K1t +   1 / [ A ]      1 / [ A2 ]     = 0.221 L /mol-s * 2.95 * 60 s   + 1 / 1.39 M          [A2 ]    =  0.0251 M        The Change in theConcentration of B depends on the Concentration of A.             From the given data :               Change in the Concentration of B = 0.00859M with repect to every 177 sec Change intime.            We can Calculate the Slope of theIInd Order Rate equation as :                        K = Y / X                              = ( 1 / [ B ] ) / t                              = 116.414 - 0.3558 / 177 - 0                               =   0.65569 L / mol-s.                T 1/2    = 1 / K [A ]0                            =  1 / 0.221 L / mol-s x 1.39 M                            =   3.255 S.      1 / [ A2]   = K1t +   1 / [ A ]      1 / [ A2 ]     = 0.221 L /mol-s * 2.95 * 60 s   + 1 / 1.39 M          [A2 ]    =  0.0251 M        The Change in theConcentration of B depends on the Concentration of A.             From the given data :               Change in the Concentration of B = 0.00859M with repect to every 177 sec Change intime.            We can Calculate the Slope of theIInd Order Rate equation as :                        K = Y / X                              = ( 1 / [ B ] ) / t                              = 116.414 - 0.3558 / 177 - 0                               =   0.65569 L / mol-s.                T 1/2    = 1 / K [A ]0                            =  1 / 0.221 L / mol-s x 1.39 M                            =   3.255 S.                            =  1 / 0.221 L / mol-s x 1.39 M                            =   3.255 S.

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