Answer:
a) [tex]k = \frac{ln\frac{C_0}{C_t} }{t} \\k = \frac{2.3}{t} log \frac{C_0}{C_t}[/tex]
b) [tex]6.31 * 10^{-4}[/tex]
c) [tex]1.094 * 10^{-3}[/tex]
Explanation:
Complete question
(min) 0 135 342 683 1693
c {M} 2.08 1.91 1.67 1.35 0.57
Solution -
a) One this clear from the data provided and it is that rate of the reaction does not remain constant with time and hence it is not a zero order reaction. Thus, the equation for first order reaction is
[tex]k = \frac{ln\frac{C_0}{C_t} }{t} \\k = \frac{2.3}{t} log \frac{C_0}{C_t}[/tex]
b) Rate of reaction
Substituting the given values in above equation, we get -
[tex]k = \frac{2.3}{135} log \frac{2.08}{1.91} \\= 6.31 * 10^{-4}[/tex]
c) Value of the half-life for the reaction in minutes
[tex]t_{\frac{1}{2}} = \frac{0.69}{k}[/tex]
Substituting the given values in above equation, we get -
[tex]t_{\frac{1}{2}} = \frac{0.69}{6.31 * 10^{-31}} \\t_{\frac{1}{2}} = 1.094 * 10^{-3}[/tex]
per minute