# Which of following statements is incorrect for a first order reaction, where $T_x$ is the time taken for the reactant to drop by that % number.

$\begin{array}{1 1}(a)\;T_{99.9\%}=10T_{50\%}\\(b)\;T_{75\%}=2T_{50\%}\\(c)\;T_{99.9\%}=25T_{25\%}\\(d)\;T_{50\%}=5\normalsize T_{25\%}\end{array}$

Answer: $T_{50\%}=5T_{25\%}$ is incorrect
$kt_{x\%} = ln\;\large\frac{A_t}{A_0}$
$\Rightarrow kt_{99.9\%} = ln \;\large\frac{1}{1-.999} $$= ln \;1000 \approx 6.9 \Rightarrow 10 kt_{50\%} = 10\;\ln\;2 \approx 6.9 \Rightarrow kt_{75\%}=ln \;\large\frac{1}{1-0.75}$$ = ln \; 4 =1.38$
$\Rightarrow 2kt_{50\%}= 2 \times 0.69 = 1.38$
$\Rightarrow kt_{99.9\%} = ln \;\large\frac{1}{1-.999} $$= ln \;1000 \approx 6.9 \Rightarrow 25 kt_{25\%} = 25\;ln \; \large\frac{1}{1-0.25}$$ =25\; ln \; \large\frac{4}{3}$$\approx 7 \Rightarrow kt_{50\%}= 0.69 \Rightarrow 5 kt_{25\%}$$ = 5 \;ln \; \large\frac{1}{1-0.25}$$=5 \; ln \; \large\frac{4}{3}$$ \approx 1.4$
Therefore, this equation is incorrect.
edited Jul 24, 2014