HW 2 Solution 2015

# HW 2 Solution 2015 - CHE4473 Kinetics Spring 2015 Homework...

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Unformatted text preview: CHE4473 Kinetics Spring 2015 Homework N°2: Rate laws, reaction order, activation energy Due Tuesday JAN 20th 2015 ___________________________________________________________________________________________________ 1. For a gas reaction at 400 K the rate is reported as: (a) What are the units of the rate constant? (b) What is the value of the rate constant for this reaction if the rate equation is expressed as ___________________________________________________________________________________________________ 2. The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 650°C than at 500°C? Use R = 8.3144 J/mol K ___________________________________________________________________________________________________ 3. In the mid-­‐nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oklahoma ants, we find Running speed, (meters/hr) Temperature, (°C) 150 13 160 16 230 22 295 24 370 28 What activation energy represents this variation in bustliness with temperature? ___________________________________________________________________________________________________ 4. The maximum allowable temperature for a reactor is 800 K. At present our operating set point is 780 K, the 20-­‐K margin of safety to account for fluctuating feed, sluggish controls, etc. Now, with a more sophisticated control system we would be able to raise our set point to 792 K with the same margin of safety that we now have. By how much can the reaction rate, hence, production rate, be raised by this change if the reaction taking place in the reactor has an activation energy of 175 kJ/mol? ___________________________________________________________________________________________________ 5. A constant volume reaction AèP is being carried out a. If a 1000 liter CSTR is used with CA0 = 2 M and inlet flow rate of 10 L/min, what reaction rate (-­‐rA) is required to achieve 70 % conversion? What is the concentration of A in the reactor? b. What rate would be required to achieve 95 % conversion in the CSTR? c. For the case of 70 % conversion of A, if the reaction is first order in A, what is the reaction rate coefficient k (include units)? And if the reaction is 2nd order in A? And if the reaction is ½ order in A? d. Using the value of the first-­‐order coefficient k found in (c), if a 1000 L PFR were used instead of the CSTR, what would be the conversion? In your own words, explain why this conversion is greater than 70 % e. Using the value of the first-­‐order coefficient k found in (c) if two CSTR’s in series (each with a volume of 500 L) were used instead of one of 1000 L, what would be the conversion? In your own words, explain why this conversion is greater than 70 % ChE 4473 Homework 2 Solution 1. a) − Spring 2015 ! = 3.66!! ℎ This is a second-­‐order rate expression in terms of partial pressure, where 3.66 = k. − !"! !" Equating units: = !"# !! and ! = = ? × ! ℎ So, = . b) Now the rate expression is expressed in terms of concentration, which means we need to convert the previous expression and its associated k to the new units. Assume ideal gas: ! = ! or ! = Sub into previous expression: !! !" ! = ! . ! = 3.66 ! ! − 1 ! = 3.66!! = −! LHS is now the rate in terms of concentration, now we just need to evaluate the new k, which is equal to 3.66. 1 ! ℎ = 3.66 400 8.205 × 10!! 3600!! = . × ! ℎ Must be in these units in order to match with the given expression, and units should be clearly indicated. 2. = 300 Arrhenius equation: ! = ! !!" ! We need to find the ratio of !! when ! = 650 ° and ! = 500 °. ! ! ! ! ! !!! ! ! = ! = ! ! !!! ! ! ! !! !! Plug in values; be sure to use absolute temperature (Kelvin). = , i.e. at 650 C the rate is nearly 2000 times faster than at 500. 3. The data shows a rate increase with temperature, indicating that there is an energy of activation. We need to generate an Excel plot of the given data in order to evaluate this activation energy. Plot ln () versus 1/, and the slope will be equal to – /, which is then evaluated to find . Make sure to use absolute temperature. The activation energy is equal to 10.8 /. Other units are acceptable, depending on choice of gas constant but must clearly indicate units in answer. 4. This problem is similar to problem number 2, with ! = 792 and ! = 780 . ! ! ! ! !!! ! ! = ! = ! ! !!! ! ! ! !! !! = . , i.e. the rate will increase by approximately 50% due to the 12 K increase. 5. a) We have a 1000 L CSTR with !! = 2.0 , ! = 10 /. 70% Conversion: ! = !! 1 − ! = 2.0 1 − 0.7 = . = Conduct a mole balance on A: = − + 0 = ! !! − ! + ! − − = = . By stoichiometry or mole balance, ! = 1.4 b) To achieve 95% conversion, ! = 0.1 Then, − − = = . c) If we have a first order reaction A, −! = ! !"# −! = 0.014 ! !"# , ! = 0.6 . So, = . =. ! Second order, = . ra = kCa^2 ra = kCa^0.5 Half order, = . /( ) d) For the PFR, the mole balance for a constant density system is given by: !!! = ! = −! , using the first-­‐order rate constant = 0.023 min!! . !" The molar flow rate is equal to the product of the steady-­‐state volumetric flow through the reactor and the concentration, ! = ! ! . ! ! = −! Separate variables and integrate: !! ! ! =− ! ! !!! ! !. = . , = = . , much higher conversion. f) Now, considering a case with two consecutive 500 L CSTRS and the first-­‐order rate expression determined previously: For the first reactor: 0 = ! !! − !! + ! Plug in rate expression ! = −!! and solve for !! , which is the concentration leaving reactor 1. !! = 0.9231 For the second reactor: 0 = ! !! − !! + ! Plug in rate expression ! = −!! and solve for !! , which is the concentration leaving reactor 2. !. = . , = = . , so a somewhat higher conversion is achieved by splitting the volume between two reactors. ...
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