Question

The reaction $A\left(aq\right)\to B\left(aq\right)+C\left(aq\right)$ is monitored by measuring optical rotation of reaction mixture at different time interval. The species $A,B$ and $C$ are optically active with specific rotations ${20}^0,{30}^0$ and $-{40}^0$ respectively. Starting with pure $A$ if the value of optical rotation was found to be ${2.5}^0$ after $6.93$ minutes and optical rotation was $-5^0$ after infinite time. Find the rate constant for first order conversion of $A$ into $B$ and $C$.

Answer 1

moles $A\left(aq\right)⟶B\left(aq\right)+C\left(aq\right)$ Rotation

at $t=0$ $90$ - - $r_o=r_A={20}^o$

at $t=6.93$ min $90-x$ $x$ $x$ $r_t={2.5}^o$

at $t=\infty$ - $90$ $90$ $r_{\infty }=-5^o$

as, $x\propto r_o-r_t$ $(a_o-(a_o-x)\propto r_o-r_t)$

$a_o\propto r_o-r_{\infty }{r}_o-r_{\infty }$ $(a_O-O\propto r_o-r_{\infty })$

$K=\frac{2.303}{t}\log \frac{a_o}{a_o-x}$

$\Rightarrow K=\frac{2.303}{6.93}\log \frac{r_o-r_{\infty }}{(r_o-r_{\infty })-(r_o-r_t)}$

$\Rightarrow K=\frac{2.303}{6.93}\log \frac{r_o-r_{\infty }}{r_t-r_{\infty }}$

$\Rightarrow K=\frac{2.303}{6.93}\log \frac{20-(-5)}{2.5-(-5)}$

$\Rightarrow K=\frac{2.303}{6.93}\log \frac{25}{7.5}$

$\Rightarrow K=0.173mi{n}^{-1}\Rightarrow$ Rate constant