If the decomposition reaction $A(g) \longrightarrow B(g)$ follows first order kinetics, then the graph of…
If the decomposition reaction $A(g) \longrightarrow B(g)$ follows first order kinetics, then the graph of rate of formation of $B$, denoted by $R$, against time $t$ will be
None of these
Solution
According to first order kinetics
$\begin{array}{lcc} & A(g) \longrightarrow B(g) \\ \text { at } t=0 & a_0 & - \\ \text { at } t=t & a_0-x & x\end{array}$
Formula for first order reaction, $A_t=A_0 e^{-k t}$
Here, $A_t=$ concentration at time ' $t$ '
$
\begin{gathered}
A_0=\text { initial concentration } \\
\therefore \quad A_0-x=A_0 e^{-k t} \\
\quad A_0\left(1-e^{-k t}\right)=x
\end{gathered}
$
$\therefore$ For $[B]_t=x$
$
\begin{aligned}
& {[B]_t=A_0\left(1-e^{-k t}\right)} \\
& {[B]_t=A_0-A_0 e^{-k t}}...(i)
\end{aligned}
$
On comparing the above equation with
y = c + mx
y = rate (R), x = time (t)
Slope, $m=1-e^{-k t}$
(i.e. exponentially decrease)
$\therefore$ Graph is
