The half-life of the reaction is 50 minutes
Data;
Let the initial concentration of the reaction be X[tex]_0[/tex]
The reactant left = (1 - 0.45) X[tex]_0[/tex] = 0.55 X[tex]_0[/tex] = X
For a first order reaction
[tex]\ln(\frac{x}{x_o}) = -kt\\ k = \frac{1}{t}\ln (\frac{x_o}{x}) \\ k = \frac{1}{43}\ln (\frac{x_o}{0.55_o})\\ k = 0.013903 min^-^1[/tex]
The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.
[tex]x = \frac{x_o}{2} \\t = t_\frac{1}{2} \\t_\frac{1}{2} = \frac{1}{k}\ln(\frac{x_o}{x_o/2})\\[/tex]
Substitute the values
[tex]t_\frac{1}{2} = \frac{1}{k}\ln(2)=\frac{0.6931}{0.013903}\\t_\frac{1}{2}= 49.85 min = 50 min[/tex]
The half-life of the reaction is 50 minutes
Learn more on half-life of a first order reaction here;
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