At $603\;K$ the reaction
$2NO_2(g)$ $\longrightarrow$ $ 2NO(g)$ $+$ $O_2$
is second order with
$k$ $=$ $0.755\frac{L}{mol\;s}$
We start with
$[NO_2(g)]_o$ $=$ $0.00650\; M$
a) What is the molarity of $NO_2(g)$ after $125\;s$ ?
b) After what time the molarity of $NO_2(g)$ will it be $0.00100\;M$ ?
c) Calculate the half-life time $\theta$ !
Let's call $NO_2(g)$ $=$ $A$
a)
For $t=125\;s$ we have:
$\frac{1}{[A]_t}=\frac{1}{[A]_o}+kt$
= $ 248$
$[A]_t$ $=$ $0.00403\;M$
b)
$t$ $=$ $\frac{1}{k}(\frac{1}{[A]_t}$ $-$ $\frac{1}{[A]_o})$
= $1.12\cdot 10^3\;s$
c)
Half-life time $\theta$ :
$[A]_\theta$ $=$ $0.00325\;M$
$\theta$ $=$ $\frac{1}{k}(\frac{1}{[A]_\theta}$ $-$ $\frac{1}{[A]_o})$
= $204\;s$
