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Deffense [45]
3 years ago
8

For the reactions below, describe the reactor system and conditions you suggest to maximize the selectivity to make the desired

species, D, which is a very small molecule. For full credit, support your suggestions with clear calculations, explanations and sketches where appropriate. Rates are in mol/L.s and concentrations are in mol/L. Reaction 1: A+B  D -r1A= 10 exp[-8000 K/T]CACB Reaction 2: B+D  U -r2B= 109 exp[-10,000 K/T]CBCD BELOW – make sure to note highlighted question number for answers on Exam Paper. 4. Describe the reactor system and conditions: ___ Answer written on Exam paper AND in BB
Chemistry
1 answer:
alexandr1967 [171]3 years ago
3 0

Answer:

hello your question lacks the required reaction pairs below are the missing pairs

Reaction system 1 :

A + B ⇒ D  -r_{1A}  = 10exp[-8000K/T]C_{A}C_{B}

A + B ⇒ U -r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}

Reaction system 2

A + B ⇒ D  -r_{1A} = 10exp( -1000K/T)C_{A}C_{B}

B + D ⇒ U  -r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}

Answer : reaction 1 : description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

reaction 2 :

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

Explanation:

Reaction system 1 :

A + B ⇒ D  -r_{1A}  = 10exp[-8000K/T]C_{A}C_{B}

A + B ⇒ U -r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}

the selectivity of D is represented  using the relationship below

S_{DU} = \frac{-r1A}{-r2A}

hence SDu = 1/10 * \frac{exp(-800K/T)}{exp(-1000K/T)} * C_{A} ^{0.5} C_{B} ^{-0.5}

description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

Reaction system 2

A + B ⇒ D  -r_{1A} = 10exp( -1000K/T)C_{A}C_{B}

B + D ⇒ U  -r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}

selectivity of D

S_{DU} = \frac{-r1A}{-r2A}

hence Sdu = 1/10^7  *  \frac{exp(-1000K/T)}{exp(-10000K/T)} *\frac{C_{A} }{C_{D} }

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

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An original sample of the radioisotope fluorine-21 had a mass of 80.0 milligrams. Only 20.0 milligrams of this original sample r
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<u>Answer:</u> The correct answer is Option 3.

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All the radioisotope decay processes follow first order kinetics.

Rate law expression for first order kinetics is given by the equation:

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Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             Q_{1} = mC_{1} \Delta T_{1}

Putting the given values into the above equation as follows.

           Q_{1} = mC_{1} \Delta T_{1}

                      = 27.3 g \times 1.70 J/g K \times 41

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           Q_{2} = energy required = mL_{v}

    L_{v} = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             Q_{2} = mL_{v}

                         = 27.3 \times 444

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Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            Q_{3} = mC_{2} \Delta T_{2}

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Hence, putting the given values into the above formula as follows.

             Q_{3} = mC_{2} \Delta T_{2}

                       = 27.3 g \times 1.06 J/g \times 34 K

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Therefore, net heat required will be calculated as follows.

            Q = Q_{1} + Q_{2} + Q_{3}

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

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