Question

Following is the graph between log $T_{50}$ and log a (a = initial concentration) for a given reaction at ${27}^0$C. Hence order is :

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

As we know,
$t_{1/2}\infty {\left(\frac{1}{a}\right)}^{n-1}or{t}_{1/2}=k(a{)}^{1-n}$
log $t_{1/2}$ = log k + (1 - n) log a (y = c + mx)
Slope = (1 - n) = tan 45 = 1
$\therefore (1-n)=1\Rightarrow n=0$