A in a global single step mechanism for butane combustion...

In a global, single-step mechanism for butane combustion, the reaction order with respect to butane is 0.15 and the reaction order with respect to oxygen is 1.6. The rate coefficient can be expressed in Arrhenius form: the pre-exponential factor A is 4.16E09 (in SI units), and the activation energy EA is 125,000 kJ/kmol; the temperature exponent is equal to zero. What are the units of A? Write an expression for the rate of butane destruction, d[C4H10]/dt. ANSWER FOR A d[C4H10] / dt = 2.08 x 10^9 e^(-1.503 x 10^5 / T) * [C4H10]^(-0.75) b) Using the results of part a, determine the numerical value of the volumetric mass oxidation rate of butane (in kg/m3-s) for a fuel-air mixture with an equivalence ratio of 0.9, temperature of 1200 K, and pressure of 1 atm.

Transcript

00:02 Welcome here in this question we have in a global single strip mechanism for butane combustion so butane combustion means a c4 h10 so this is a butane which is combustive with water oxygen the reaction is co2 and h2o so as it is formos these four modes has it is 10 most this one is five most and here five and eight 13 are present so 13 divided by 2 so this is a reaction here so if you see they are given that the order of the reaction with respect to butin is 1 .5 and with respect to oxygen it is 1 .6 it is given so here we need to find out the reaction activation energy is given so activation energy is one two five zero zero kilo zero kilos per mole and we have the pre -exponential factor four point one six multiplied by ten per mole kilo mole kilo per meter cube whole power zero minus zero point seven five is given so pre -exponential factor and the activation energy is given so we need to find out the expression for c4 h10 divided by dd value and we need to find out the volumetric mass oxidation rate of butane we need to find out here so we will see here the rate of this one is equal to k multiplied by c4 h 10 and o2 concentration here it is 0 .15 and this one is 1 .6 and k is equal to arrainius equation ae power exponential power a minus a minus a divided by r t so a value is given 4 .16 multiplied by 10 power 9 minus 0 .75 it is given and here the value is exponential power 1.

02:46 This one is 1 2 5000 divided by 8 .314 multiplied by 10 power so multiplied by t this value here so from this we can write here the k value is approximately it is a d c4 h10 is equal to divided by d t is equal to 0 .15 k c4 h10 power 0 .15 multiplied by oxygen 1 .6 this is the value so this is the value of this one and now we will write volumetric oxidation rate of butane we can write here minus row d .c4h10 divided by dt is the formula here.

03:50 So as we need to find out the density.

03:53 So this is the density of the mixer.

03:58 So density of the mixer, we will find out given an.

04:03 Equivalence ratio is given 0 .9 % of butane is given.

04:10 So molar ratio of molar ratio of butin is 1 divided by 13 .6 so assuming 80s nitrogen is 72 % and oxygen is 21 % by volume.

04:31 So we can write calculate the molar masses on the concentrations of the species in the mixture.

04:40 So m, c4h10 is equal to 58 .1 2 gram per mole.

04:48 Then concentration of c4h10 is equal to 0 .9 divided by 13 .6 row multiplied, so here it is multiplied by, so divided by 58 .12.

05:07 Then it is equal to approximately this value is this much here we will have.

05:17 Now we will write the molecular mass of o2 is 32 grams and concentration of o2 is equal to 0 .0.

05:32 066 divided by p divided by 32 and here it is mn2 that is equal to 28 and concentration of n2 is equal to 0 .79 divided by 0 .21 and concentration of what we will get here is 3 .754 approximately.

06:13 Now the total number of most total molar mass of the mixture is equal to so for ch4 and concentration of ch4 we can write n i m i we can write so here the total is equal to 0 .9 divided by 13 .6 r.

06:40 Plus 0 .66 multiplied by 3 .754 r...

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