Question

#1 when table sugar (sucrose), C12H22O11(s) is digested, it is broken down by reacting with water to produce two simple sugars, glucose and fructose, each with the formula C6H12O6(s).

C12H22O11(s) + H2O(l) -----> C6H12O6(s) + C6H12O6(s)

An experiment was performed, and the following data were obtained ;( 4marks)

Run	initial (sucrose) (mol/L)	Rate (mol/(Ls))
1	0.050	3.09x10^-5
2	0.100	6.17x10^-5
3	0.250	1.50x10^-4

#2 For the reaction A + B +2C --> ABC2, the following data were obtained:

run	(a) (mol/L)	(B) (Mol/L)	(C) (mol/L)	Rate (mol/Ls))
1	0.120	o.220	1.25	2.14x10^-3
2	0.120	0.220	2.50	2.14x10^-3
3	0.120	0.440	1.25	4.28x10^-3
4	0.240	0.220	1.25	8.56x10^-3

A) Determine the value of the rate law ( 4 marks )

B) Determine the rate law for this reaction ( 4 marks )

Answer 1

1) C₁₂H₂₂O_11(s) + H₂O_{(l) -}----> C₆H₁₂O_6(s) + C₆H₁₂O_6(s)

Let the rate law is given by, Rate = k [C₁₂H₂₂O₁₁]^m k =rate constant

Given that, Rate =3.09x10^-5mol/ls , [C₁₂H₂₂O₁₁]= 0.050 mol/L -------------- (1)

Rate =6.17 x 10^-5mol/ls , [C₁₂H₂₂O₁₁]= 0.100 mol/L -------------- (2)

equation (2) / equation (1) ,(applying rate law)

(6.17 x 10^-5) / (3.09x10^-5) = (0.100 / 0.050)^m

or, 2 = 2^m

or, m = 1 i.e.- order w.r.t. C₁₂H₂₂O₁₁ is 1 i.e.- 1^st order

Thus the rate law is, Rate =k[C₁₂H₂₂O₁₁]

let, Rate =3.09x10^-5mol/ls , [C₁₂H₂₂O₁₁]= 0.050 mol/L

k = Rate / [C₁₂H₂₂O₁₁] = 3.09x10^-5 mol/ls / 0.050 mol/L

or, k = 6.18 x 10^-4 s^-1