The following data are for questions through F and refer to the reaction: A+2B+3

Question

The following data are for questions through F and refer to the reaction:

A+2B+3C--> 2Y + Z   All data were taken at 50.0degrees celcius

Trial     Initial [A]    Initial [B]    Initial [C]    Rate of [Y]

1           0.10            0.02           0.04            10M/s

2            0.10           0.03           0.04            15M/s

3           0.20            0.02           0.08            80M/s

4           0.20            0.02           0.16            160M/s

5           0.05            0.01           0.08                ?

A.) what is the rate of formation of Y if [B] is doubled?

B.) What is the rate of formation of Z in trial 3 (in M/s)?

C.) What is the rate of disappearance of C in trial 2 (in M/s)?

D.) What is the rate law derived for the above data?

E.) What is the missing rate (trial 5) in M/s?

F.) What is the rate constant?

Please show working

Explanation / Answer

a) trial 1 & trial 2

let r = K Ca^a * Cb^b * Cc^c

trial 1 , 10 = K (0.1)^a * (0.02)^b * (0.04)^c

trial 2 , 15 = K (0.1)^a * (0.03)^b * (0.04)^c

divinding (1) by (2) ,

10/15 = ( 0.02/0.03) ^b

b = 1

r = K Ca^a * Cb^1 * Cc^c = K Ca^a * Cb * Cc^c

so if [B] is doubled , rate formation = K Ca^a * 2*Cb* Cc^c = 2* K Ca^a * Cb * Cc^c

so the rate of formation of Y is doubled if [B] is doubled

b) rate of formation of Y/2 = rate of formation of Z

so, in trial 3 rate of formation of Z = 80/2 = 40 M/s

c) rate of disappearance of C in trial 2 (in M/s)/3 = rate of formation of Y/2

rate of disappearance of C in trial 2 (in M/s) = (3/2) * rate of formation of Y = 1.5 * 15 = 22.5 M/s

d) trial 3 , r = 80 = K (0.2)^a * (0.02) * (0.08)^c

trial 4 , r = 160 = K (0.2)^a * (0.02) * (0.16)^c

dividind 3 by 4,

80/160 = (0.08/0.16)^c

c = 1

so r = K Ca^a * Cb * Cc

trial 2 , 15 = K (0.1)^a * (0.03)* (0.04)

trial 3 , 80 = K (0.2)^a * (0.02) * (0.08)

dividing 2 by 3 ,

15/80 = (0.03/0.02) * (0.04/0.08)*(0.1/0.2)^a

a = 2

so, r = K Ca^2 * Cb * Cc

trial 4 , r = 160 = K (0.2)^2 * (0.02) * (0.16)

K = 1250000

so, r = 1250000* Ca^2 * Cb * Cc is the rate equation.

e) trial 5 , r = 1250000* 0.05^2 * 0.01 * 0.08 = 2.5 M/s

missing rate (trial 5) in M/s = 2.5 M/s

f) rate constant = K = 1250000

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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