Recombination of iodine atoms in the gas phase in the presence of argon obeys the equation:
$2I(g)$ $+$ $Ar(g)$ $\longrightarrow$ $I_2(g)$ $+$ $Ar(g)$
Here experimental results for the rate of formation of $I_2$ (in $\frac{mol}{L\cdot s})$ depending on the initial concentrations of the reagents (in $\frac{mol}{L})$:
$[I(g)]$
$[Ar(g)]$
$v_{I_2}$
$1.0\cdot 10^{-5}$
$1.0\cdot 10^{-3}$
$8.7\cdot 10^{-4}$
$2.0\cdot 10^{-5}$
$1.0\cdot 10^{-3}$
$3.48\cdot 10^{-3}$
$2.0\cdot 10^{-5}$
$5.0\cdot 10^{-3}$
$1.74\cdot 10^{-2}$
Find the rate law! What about the orders of reagents?
- As the coefficient of $I_2$ is $1$, we have:
$v_r$ $=$ $\frac{\Delta [I_2]}{1\cdot \Delta t}$ $=$ $\frac{\Delta [I_2]}{\Delta t}$ $=$ $v_{I_2}$
- The rate law
$v_r=k[I(g)]^{\alpha}[Ar]^{\beta}$
becomes:
(1) $8.7\cdot 10^{-4}$ $=$ $k(1.0\cdot 10^{-5})^{\alpha}(1.0\cdot 10^{-3})^{\beta}$
(2) $3.48\cdot 10^{-3}$ $=$ $k(2.0\cdot 10^{-5})^{\alpha}(1.0\cdot 10^{-3})^{\beta}$
(3) $1.74\cdot 10^{-2}$ $=$ $k(2.0\cdot 10^{-5})^{\alpha}(5.0\cdot 10^{-3})^{\beta}$
Dividing (2) by (1):
$4 = 2^{\alpha}$
$\alpha=2$
Dividing (3) by (2):
$5 = 5^{\beta} $
$\beta=1$
Introducing to (1):
$k=$ $\frac{8.7\cdot 10^{-4}}{(1.0\cdot 10^{-5})^2\cdot 1.0\cdot 10^{-3}}$ $ = $ $8.7\cdot 10^9 \frac{L^2}{mol^2\cdot s}$
Rate law:
$v_r$ $=$ $8.7\cdot 10^9[I(g)]^2[Ar(g)]$
Order is (here) always equal to molecularity!
