For the reaction A + B + C D The following data were obtained at constant temper

Question

For the reaction
A + B + C D
The following data were obtained at constant temperature:

Part A) What is the order with respect to each reactant? (Enter the order for A as Answer 1, that for B as Answer 2, and that for C as Answer 3. You must get all 3 correct.)

Part B) What is the value of the rate constant for the reaction at this temperature?

Trial Initial [A] (mol/L) Initial [B] (mol/L) Initial [C] (mol/L) Initial Rate (mol/L . s) 1 0.4905 0.4215 0.0843 11.416 2 0.9810 0.4215 0.0843 22.832 3 0.9810 0.8430 0.0843 91.326 4 0.4905 0.4215 0.1686 11.416

Explanation / Answer

rate = k[A]^m[B]^n [C]^p

by observing trial1 and trial2

11.416 = K[0.4905]^m[0.4215]^n[0.0843]^p    ...........1

22.832 = K[0.9810]^m[0.4215]^n[0.0843]^p ........2

do eq1/eq2

11.416/22.832 =[0.4905/0.9810]^m

( 1/2)^1 = [1/2]^m

bases are equal powers should be equated

so m= 1

order w.r.t A is 1

by observing trial1 and trial4

11.416 = K[0.4905]^m[0.4215]^n[0.1686]^p .................3

do eq1/eq2

1 = [0.0843/0.1686]^p

1 = (1/2)^p

(1/2)^0 = (1/2)^p

so p = 0

oredr w.r.t. C is 0

by observing trial1 and trial 4

91.326 = K[0.9810]^m [0.8430]^n[0.0843]^p ....................4

do eq1/eq2

11.416/91.326 = [0.4905/0.9810]^m[0.4215/0.8430]^n [0.0843/0.0843]^p

    1/8 = (1/2)^m(1/2)^n

we know m =1

(1/2)^3 = (1/2)(1/2)^n

(1/2)^2 = (1/2)^n

n = 2

so order w.r.t B is 2

total order of the reaction is = (1+2+0) = 3

....................................................................................

rate = K[A][B]^2

11.416 = K[0.4905][0.4215]^2

K   = 1.31*10^2 M^-2 sec^-1

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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