Question

t (s)	[A] (M)	ln[A]	1/[A]
0.00	0.500	−0.693	2.00
20.0	0.389	−0.944	2.57
40.0	0.303	−1.19	3.30
60.0	0.236	−1.44	4.24
80.0	0.184	−1.69	5.43

Now that we know it os first order

What is the value of the rate constant for this reaction?

Answer 1

Tabulate the data as follows

Time (s)                                              0                   20                    40      60               80

Concentration of A in M, [A]               0.500            0..389               0.303               0.236          0184

For a first order reaction the rate constant k is given by the equation

    k = 1/t ln C₀/C_t

where C₀ = Initial concentration ie concentration at time 0

        C_t = concentration at time t

When t = 0 s         C₀ = 0.500 M

When t = 20 s       C_t = 0..389 M    k = 1/20   ln 0.500 M/ 0..389 M = 0.0125 s^-    

When t = 40 s       C_t = 0.303 M    k = 1/40   ln 0.500 M/ 0.303 M = 0.0125 s^-   

When t = 60 s       C_t = 0.236 M    k = 1/60   ln 0.500 M/ 0.236 M = 0.0125 s^-   

When t = 80 s       C_t = 0.184 M    k = 1/80   ln 0.500 M/ 0.184 M = 0.0125 s^-   

The average of the k values is 0.0125 s^-   

The rate constant for this reaction = 0.0125 s^-