Question

1. Using data Irom Part A determine the value oly re in y- -1.322. or 2. Using data from Part B determine the value of x. e us 3. Using the values of x and y, determine the values of the rate constants for experiments A2 and B3 K-Rate k (A2)- k (B3) -뉴 Using the values of x and y, determine the values of the rate constants for experiments C1, C2, C3 and C4. 4. k(CI)= k (C2) k (C3) = k (C4) = Using these k values, plot a graph of In k vs 1/T, and determine the activation energy, E, and the value of A. Value of A = 5. Using the values of x and y, determine the values of the rate constants for experiments E1, E2, E3 and E4. k (El)- k (E2) = k (E3) = k (E4)- Using these k values, plot a graph of In k vs 1/T, and determine the activation energy E, for the catalyzed reaction, and the value of A, for the catalyzed reaction E, (catalyzed) = Value of A =

HELP! I'm not sure how to find the rate constants

Answer 1

Reaction Rate =K* [(NH₄)₂S₂O₈]^x*[KI]^y -(1)

Given, x=-1.12, y=-1.32

3), For A2 reaction, [(NH₄)₂S₂O₈]=0.2 mol/litre, [KI]=0.2 mol/litre

Given, Reaction rate = 0.004 mol/litre/second

Putting values in (1), K(A2) = 8.13*10^-5 mol^3.44 litre^-3.44 s

For B3 reaction, [(NH₄)₂S₂O₈]=0.1 mol/litre, [KI]=0.2 mol/litre

Given, Reaction rate = 0.009 mol/litre/second

Putting values in (1), K(B3) = 8.42*10^-5 mol^3.44 litre^-3.44 s

4)

For C1 reaction, [(NH₄)₂S₂O₈]=0.2 mol/litre, [KI]=0.1 mol/litre

Given, Reaction rate = 0.004 mol/litre/second

Putting values in (1), K(C1) = 3.3*10^-5 mol^3.44 litre^-3.44 s

For C2 reaction, [(NH₄)₂S₂O₈]=0.2 mol/litre, [KI]=0.1 mol/litre

Given, Reaction rate = 0.002 mol/litre/second

Putting values in (1), K(C2) = 1.65*10^-5 mol^3.44 litre^-3.44 s

For C3 reaction, [(NH₄)₂S₂O₈]=0.2 mol/litre, [KI]=0.1 mol/litre

Given, Reaction rate = 0.0004 mol/litre/second

Putting values in (1), K(C3) = 0.33*10^-5 mol^3.44 litre^-3.44 s

For C4 reaction, [(NH₄)₂S₂O₈]=0.2 mol/litre, [KI]=0.1 mol/litre

Given, Reaction rate = 0.0003 mol/litre/second

Putting values in (1), K(C4) = 0.24*10^-5 mol^3.44 litre^-3.44 s

Reaction	K	T, degC	T, deg K	lnK	1/T
C1	3.3*10^-5	40	313	-4.45	.0031
C2	1.65*10^-5	23	296	-4.82	.0033
C3	0.33*10^-5	13	286	-5.5	.0034
C4	0.24*10^-5	4	277	-5.63	.0036

Performing regression on above values, log(K) = 3.378+(-2531)*R/T

General equation is, log(K) = log(A)-Ea/R/T

Comparing both, log(A)=3.378, => A = 2344.2 mol^3.44 litre^-3.44 s

-Ea/R = -2531

R=8.3 joule/mol/K

Therefore E=21 KJ/mol

5)

For E1 reaction, [(NH₄)₂S₂O₈]=0.2 mol/litre, [KI]=0.1 mol/litre

Given, Reaction rate = 0.02 mol/litre/second

Putting values in (1), K(E1) = 0.00016 mol^3.44 litre^-3.44 s

For E2 reaction, [(NH₄)₂S₂O₈]=0.2 mol/litre, [KI]=0.1 mol/litre

Given, Reaction rate = 0.01 mol/litre/second

Putting values in (1), K(E2) = 8.26*10^-5 mol^3.44 litre^-3.44 s

For E3 reaction, [(NH₄)₂S₂O₈]=0.2 mol/litre, [KI]=0.1 mol/litre

Given, Reaction rate = 0.001 mol/litre/second

Putting values in (1), K(E3) = 0.82*10^-5 mol^3.44 litre^-3.44 s

For E4 reaction, [(NH₄)₂S₂O₈]=0.2 mol/litre, [KI]=0.1 mol/litre

Given, Reaction rate = 0.001 mol/litre/second

Putting values in (1), K(CE4) = 0.82*10^-5 mol^3.44 litre^-3.44 s

Reaction	K	T, degC	T, deg K	lnK	1/T
E1	16*10^-5	40	313	-8.74	.0031
E2	8.26*10^-5	23	296	-9.4	.0033
E3	0.82*10^-5	13	286	-11.7	.0034
E4	0.82*10^-5	4	277	-11.7	.0036

Performing regression on above values, ln(K) = 11.65+(-6577)*R/T

General equation is, ln(K) = ln(A)-Ea/R/T

Comparing both, ln(A)=11.65, => A = 114691 mol^3.44 litre^-3.44 s

-Ea/R = -6577

R=8.3 joule/mol/K

Therefore E=54.5 KJ/mol