Question

A graph plotted between $log{t}_{50}$% vs. $log$ concentration is a straight line. What conclusion can you draw from the given graph?

• A. $n=1,t_{1/2}=\frac{1}{K\cdot a}$
• B. $n=2,t_{1/2}=1/a$
• C. $n=1,t_{1/2}=\frac{0.693}{K}$
• D. none of the above

Answer 1

Answer: (C)

$n=1,t_{1/2}=\frac{0.693}{K}$

The graph gives the values : $n=1,t_{1/2}=\frac{0.693}{K}$

$t_{1/2}\propto (a{)}^{1-n}$ or
$t_{1/2}=Z\cdot (a{)}^{1-n}$ or
$log{t}_{1/2}=logZ+(1-n)loga$ or
$\left(y\right)=c+mx$
Thus, slope $=m=1-n$ or $1-n=0\therefore n=1$ and for Ist order reaction $t_{1/2}=\frac{0.693}{K}$.