For a second-order reaction, [A]?products, the rate of the reaction is given as
ID: 812026 • Letter: F
Question
For a second-order reaction, [A]?products, the rate of the reaction is given as rate= k[A]2, where k is the rate constant and [A] is the concentration of reactant A. The integrated rate law for second-order reactions is 1[A]t=kt+1[A]0, where [A]t is the concentration of reactant A at time t, k is the rate constant, and [A]0 is the initial concentration of reactant A. This equation is of the type y=mx+b. Therefore, the plot of 1[A]t versus time is always a straight line with a slope k and a y intercept 1[A]0.
Consider the second-order reaction: 2HI(g)?H2(g)+I2(g) Use the simulation to find the initial concentration [HI]0 and the rate constant k for the reaction. What will be the concentration of HI after t = 7.311010s ([HI]t) for a reaction starting under the condition in the simulation?
For a second-order reaction, [A]?products, the rate of the reaction is given as rate= k[A]2, where k is the rate constant and [A] is the concentration of reactant A. The integrated rate law for second-order reactions is 1[A]t=kt+1[A]0, where [A]t is the concentration of reactant A at time t, k is the rate constant, and [A]0 is the initial concentration of reactant A. This equation is of the type y=mx+b. Therefore, the plot of 1[A]t versus time is always a straight line with a slope k and a y intercept 1[A]0. Consider the second-order reaction: 2HI(g)?H2(g)+I2(g) Use the simulation to find the initial concentration [HI]0 and the rate constant k for the reaction. What will be the concentration of HI after t = 7.31½1010s ([HI]t) for a reaction starting under the condition in the simulation?Explanation / Answer
Because we dont have numerical data but theoretical data, we assume a constant volume of 1 liter and initial moles of HI 1 mol.
As the reaction proceeds the following occurs:
2HI + H2 > I2
Initial: 1mol 1 mol -
Reaction: 2x x x
Final: 1-2x 1-x x
Lets say then that if we measure the concentration of HI every so often and make a comparison chart, we can plot the result and obtain the values of k and the initial concentration.
Suppose a solution of 1 liter where the initial concentration of HI is unknown and what we do is measure their concentration from 10 seconds every 5 seconds for 1 minute. Then plotted (1 / [HI]) versus time:
HI (mol/L)
t (s)
1/HI (L/mol)
0,95
10
1,0526
0,92
15
1,0870
0,9
20
1,1111
0,87
25
1,1494
0,83
30
1,2048
0,79
35
1,2658
0,76
40
1,3158
0,72
45
1,3889
0,68
50
1,4706
0,63
55
1,5873
0,6
60
1,6667
Excel plotting time on the x axis and (1/HI) on the Y axis, then do a linear fit in order to obtain the the equation:
(1/[HI]t)= 0,013 t + 0,9009
Where k= 0,013 and [HI]o= (1/0,9009) = 1.11 mol/L
In this equation the concentration on the time of 7.31 x1010 would be:
1/[HI] = (0,013*7,31x1010) + 0,9009
1/[HI] = 9,5 x 108
[HI] = 1/(9,5 x 108) = 1 x 10-9 = almost cero.
HI (mol/L)
t (s)
1/HI (L/mol)
0,95
10
1,0526
0,92
15
1,0870
0,9
20
1,1111
0,87
25
1,1494
0,83
30
1,2048
0,79
35
1,2658
0,76
40
1,3158
0,72
45
1,3889
0,68
50
1,4706
0,63
55
1,5873
0,6
60
1,6667
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