#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 )
1) C12H22O11(s) + H2O(l) -----> C6H12O6(s) + C6H12O6(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 = 2m
or, m = 1 i.e.- order w.r.t. C12H22O11 is 1 i.e.- 1st 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
#1 when table sugar (sucrose), C12H22O11(s) is digested, it is broken down by reacting with water...