Thermodynamics filled in class notes_Part_90

Thermodynamics filled in class notes_Part_90 - 5.5....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION 187 Now use Eq. (5.350) to eliminate the density derivative in Eq. (5.338) to get dρi ρ ρRT r = νi r − i dt ρ P N k =1 νk ρ RT νi = r −i P reaction effect , (5.352) νk , k =1 N (5.353) expansion effect or dρi = r dt νi reaction effect − yi ∆n expansion effect . (5.354) There are two terms dictating the rate change of species molar concentration. The first, a reaction effect, is precisely the same term that drove the isochoric reaction. The second is due to the fact that the volume can change if the number of moles change, and this induces an intrinsic change in concentration. Note that the term ρi RT /P = yi, the mole fraction. Example 5.14 Study a variant of the nitrogen dissociation problem considered in an earlier example, see p. 147, in which at t = 0 s, 1 kmole of N2 exists at P = 100 kP a and T = 6000 K . In this case, take the reaction to be isothermal and isobaric. Consider again the elementary nitrogen dissociation reaction N2 + N2 ⇌ 2 N + N2 , (5.355) which has kinetic rate parameters of a = 7.0 × 1021 β = cm3 K 1.6 , mole s −1.6, E = (5.356) (5.357) cal 224928.4 . mole (5.358) In SI units, this becomes 3 a = E = m3 K 1.6 1m 1000 mole cm3 K 1.6 = 7.0 × 1018 , mole s 100 cm kmole kmole s cal J kJ kJ 1000 mole 4.186 = 941550 224928.4 . mole cal 1000 J kmole kmole 7.0 × 1021 (5.359) (5.360) At the initial state, the material is all N2 , so PN2 = P = 100 kP a. The ideal gas law then gives at CC BY-NC-ND. 18 November 2011, J. M. Powers. 188 CHAPTER 5. THERMOCHEMISTRY OF A SINGLE REACTION t=0 P ρN 2 t=0 = PN2 = ρN2 RT, P , = RT 100 kP a , = kJ 8.314 kmole K (6000 K ) kmole = 2.00465 × 10−3 . m3 (5.361) (5.362) (5.363) (5.364) Thus, the initial volume is V |t=0 = nN2 |t=0 1 kmole = ρN2 t=0 2.00465 × 10−3 kmole m3 = 4.9884 × 102 m3 . (5.365) In this isobaric process, one always has P = 100 kP a. Now, in general P = RT (ρN2 + ρN ); (5.366) therefore, one can write ρN in terms of ρN2 : ρN = = = P − ρN 2 , RT 100 kP a − ρN2 , kJ 8.314 kmole K (6000 K ) 2.00465 × 10−3 kmole m3 (5.367) (5.368) − ρN 2 . (5.369) Then the equations for kinetics of a single isobaric isothermal reaction, Eq. (5.353) in conjunction with Eq. (5.280), reduce for N2 molar concentration to dρN2 dt = aT β exp −E RT ′ ′ (ρN2 )νN2 (ρN )νN 1− 1 (ρ )νN2 (ρN )νN Kc N2 νN2 − ρN2 RT (νN2 + νN ) . P =r (5.370) ′ ′ Realizing that νN2 = 2, νN = 0, νN2 = −1, and νN = 2, one gets dρN2 dt = aT β exp −E RT ρ2 2 1 − N 1 ρ2 N K c ρN 2 −1 − ρN2 RT P . (5.371) =k ( T ) The temperature dependency of the reaction is unchanged from the previous reaction: k (T ) = = = aT β exp −E RT 7.0 × 1018 , m3 K 1.6 kmole s 7.0 × 1018 exp T 1.6 (5.372) −1.1325 × 105 T CC BY-NC-ND. 18 November 2011, J. M. Powers. kJ kmole kJ kmole K T −941550 T −1.6 exp 8.314 . , (5.373) (5.374) ...
View Full Document

Ask a homework question - tutors are online