You are carrying out a reaction in the laboratory. You need toconvert 1.00 mole

Question

You are carrying out a reaction in the laboratory. You need toconvert 1.00 mole of reactant X into
products. This reaction shows first-order kinetics, and thereaction mixture has a constant volume
of 1.00 L.

• At 1:00pm, you start the reaction at 25°C with 1.00mole of reactant X.
• At 2:00pm, you find that 0.60 moles of reactant X remain.You immediately increase the
temperature of the reaction mixture to 35°C.
• At 3:00pm, you discover that 0.20 moles of reactant X arestill present. You want to be able to
finish the reaction by 4:00pm, but you cannot stop the reactionuntil only 0.01 moles of reactant X
remain. You must increase the temperature again.

What is the minimum temperature required to complete the reactionby 4:00pm? (Remember, the
reaction is complete when only 0.01 moles of X remain.)

Explanation / Answer

Formula :                ln (k1 /k2 ) = Ea /R ( 1/ T2 - 1/T1) Where k1 = rate constat at time T1           k2 = rate constant at time T2 Ea is the activation energy Formula for first order reaction:                ln [A]0 / [A] = kt a )   [A]0   = 1.00 mol /1.0 L                 = 1 M      [A]      = 0.60 mol / 1.0 L                 = 0.60 M       t         = 1 h       k    =1/1h * ln ( 1 / 0.60 )                = 0.510 h-1 When temperature is rised by 100c , rate isdoubled. i.e = 2 *   0.510 h-1      = 1.02 h-1 ( let thisis k1) Now by using the two we can calculate Ea . ln (   0.510 /1.02 ) = Ea / 8.314 J /mol.K ( 1/ 308 K - 1/ 298 K)                               = Ea * 10 K / 8.314J / mol.K * 308 K * 298 K                                =Ea *1.31 * 10 ^ -5 mol / J                     -0.693= Ea *1.31 * 10 ^ -5 mol / J                           Ea = -0.693 /1.31 * 10 ^ -5 mol / J                                 = -52882.28 J / mol                                = - 52.882 kJ / mol Now during the 1h from 3.00 pm to 4 pm ,    [A]0   = 0.20 M    [A]    = 0.01 M     t       =1h     k      = 1/1h ln ( 0.20 / 0.01 )            = 2.99 h-1( This is k2) Now we have k1 , k2 and T1 and Ea. From that T2 can becalculated. ln ( 1.02 / 2.99 ) = - 52.882 * 1000 J / mol / 8.314 J / mol.K( 1 / 308 K - 1/ T2 )          1.075       = 6360.59 ( 1 / 308K - 1/ T2 ) ( 1 / 308 K - 1/ T2 ) = 1.075 / 6360.59                                  =1.6 * 10 ^ -4                        1/T2 = 0.0032 - 1.6 * 10 ^ -4                                  =0.003 K-1                             T2 =329 K                                  =56.9 0C                             T2 =329 K                                  =56.9 0C

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