Question is Using either the differential or the fractional life method, determi

Question

Question is Using either the differential or the fractional life method, determine a reaction rate equation (rate law)
that represents the data in the table - report the corresponding reaction order value at the “0.1” (one decimal place)
accuracy.

I tried the Fractional method way where F=0.8 but I don't know how to find Tf, from the solution. Can someone explain?

If you don;t mind please explain to me what it is the differential method?

SOLUTION:

Problem 1 (45 pts)-Reactan1.4 decomposes in a batch reactor according to: A products. The reaction is temperature dependent and it obeys the Arrhenius law in a temperature range from 0-60°C. The concentration of A in the reactor is measured at various times and is presented in Table 1 and Figure 1 (no P was initially present in the reactor) Table 1: Time-dependent variation of concentration of A in an aqueous solution during its oxidation of A to products at 25°C. Activation energy for the reaction is 75 kJ/mol. Note: M = mol/dm3-The results graphically in Figure 1 are presented 0.5 7.3 4 3.9 10 2.8 [4]/M 10 3.2 10 4 2 0 4 6 9 10

Explanation / Answer

For a nth order reaction, the rate expression can be written as

-rA= -dCA/dt= KCAn

Which upon separating the variable and integration between t= 0 CA=CAO and t=t, CA=CA

Gives

CA(-n+1)- CAO(-n+1)=kt*(n-1)    (1)

CAO(-n+1) {CA/CAO)(-n+1)-1}= kt*(n-1)       (2)

At two different time intervals

Time t= 2 seconds CA/CAO = 0.5

Time t=1 sec CA/CAO= 0.61

Substituting these values in Eq.2 generates two equations and taking their ratio gives

{ (0.5)(-n+1)- 1}   = 2* { (0.6)(-n+1)- 1}

2* 0.6)(-n+1 } -0.5 (-n+1)= 1

Which gives only when n=1 the order of the reaction is 1

-dCA/dt= kCA

CA/CAO= e-kt

lnCA= lnCAo-kt

so a plot of lnCA vs time gives a straight line whose slope is -k

The slope is -0.112

therefore k= 0.112

The rate expression becomes -rA= 0.112CA

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.