1aa3-2012-tut6-a

1aa3-2012-tut6-a - CHEMISTRY 1AA3 Week of TUTORIAL PROBLEM...

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Unformatted text preview: CHEMISTRY 1AA3 Week of FEBRUARY 13, 2012 TUTORIAL PROBLEM SET 6 – Solutions ______________________________________________________________________________ Page 1 of 10 Kinetics questions: 1. The combination of CF 3 "free radicals", 2 CF 3 C 2 F 6 , occurs with rate constants of 5.9 10 9 M 1 s 1 at 25 C and 7.1 10 9 M 1 s 1 at 60 C. (a) Use the Arrhenius equation for the rate constant as a function of temperature to obtain the activation energy. (b) At a certain temperature the product can be condensed to a liquid. The rate constant at this temperature is 2.30 10 9 M 1 s 1 . Determine the temperature, in degrees Celsius, at which this rate constant applies. SOLUTION: (a) Using k = A e (-Ea/RT) in natural log form: ln k = -E a + lnA and substituting RT for the two temperatures, 298 K and 333 K (recall the Arrhenius relationship uses Kelvin not Celsius temperatures). ln (5.9 10 9 ) = ln A -E a . 1 --------(1) R 298 ln (7.1 10 9 ) = ln A -E a . 1 --------(2) R 333 Then (1) - (2) gives ln 5.9 10 9 = -E a . ( 1 - 1 ) 7.1 10 9 R 298 333 or ln (0.8310) = -E a (333 - 298) 8.314 J mol-1-1 298 333 or - 0.1851 = -E a 35 8.314 298 333 Isolate for E a , E a = 0.1851 8.314 298 333 35 = 4363 J mol- 1 or 4.36 kJ mol- 1 (b) If the rate constant at the cooler temperature is 2.30 10 9 M 1 s 1 , then we may use the data either from the 25 C or 60 o C experiment together with the calculated E a to obtain the value of the temperature. 2 Thus ln (5.9 10 9 ) = ln A - 4.36 kJ . 1 --------(1) R 298 and ln (2.30 10 9 ) = ln A - 4.36 kJ . 1 --------(2) R T 2 Take equation (2) -(1) ln(5.9 x 10 9 ) - ln (2.30 10 9 ) = - 4.36 10 3 J mol-1 . ( 1 - 1 ) . 1 8.314 J mol-1-1 298 T 2 K or 0.9420 = -4.36 10 3 (3.356 10 3- 1 ) 8.314 T 2 5.152 10 3 = 1 or T 2 = 194 K, and 194 – 273 = 79 C T 2 2. A reaction has an activation energy of 42.6 kJ/mol and a rate constant of 0.045 s-1 at 25.0 o C. At what temperature will the reaction rate quadruple? a. 74.2 o C b. 121.5 o C c. 51.1 o C d. 32.7 o C e. 65.3 o C SOLUTION : answer c: 51.1C Modify the Arrhenius equation shown below to resolve for T 2 . T 1 is 25 o C = 298 K; k 2 /k 1 = 4 (you don’t need to actually use the given rate constant for T 1 of 0.045 s-1 , as the question only asks at what temperature T 2 the rate will quadruple); and E a is given: T 2 = -1/((ln 4) *R/E a-1/T 1 ) = 324.1 K or 51.1 o C 2 1 a T 1 T 1 E R ln 1 2 k k 3 IMF questions: 3. How does each of the following affect the rate of evaporation of a liquid in an open dish?...
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1aa3-2012-tut6-a - CHEMISTRY 1AA3 Week of TUTORIAL PROBLEM...

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