Respuesta :
Solution :
Time (sec) Volume of NaOH (mL)
339 26.23
1242 27.80
2745 29.70
4546 3.81
[tex]$\infty$[/tex] 39.81
Now the example of the first order kinetics w.r.t volumetric analysis is :
[tex]$k=\frac{2.303}{t} \log \left(\frac{v_{\infty}-v_0}{v_{\infty}- v_t}\right)$[/tex]
Here, [tex]$v_{\infty}= \text{ volume at }\infty = 39.81$[/tex]
[tex]$v_{t}= \text{ volume at time 't' } = 27.80$[/tex]
[tex]$v_0$[/tex] = volume at time 0 = 0
Since the interval is not constant, we take the time interval as
[tex]$=\frac{903+1503+1801}{3}$[/tex]
[tex]$=\frac{4207}{3}$[/tex]
= 1402.3333
≈ 1402 seconds
[tex]$k=\frac{2.303}{1402} \log \left(\frac{39.81-0}{39.81-27.80}\right)$[/tex]
[tex]$=(0.001643) \log \left(\frac{39.81}{10.01}\right)$[/tex]
= 0.001643 x 0.52045
= 0.00082
[tex]$= 8.55 \times 10^{-4} \ sec^{-1}$[/tex]
Therefore, the first order rate constant is k [tex]$= 8.55 \times 10^{-4} \ sec^{-1}$[/tex].
