Because I (iodide) is better leaving group than Cl, so it will leave when this molecule is reacted with strong base (sodium tert-butyl oxide) giving the elimination product provided in picture<span />
Answer:
The rate law for second order unimolecular irreversible reaction is
![\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20k.t%20%2B%20%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D)
Explanation:
A second order unimolecular irreversible reaction is
2A → B
Thus the rate of the reaction is
![v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}](https://tex.z-dn.net/?f=v%20%3D%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D%20k.%5BA%5D%5E%7B2%7D)
rearranging the ecuation
![-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Integrating between times 0 to <em>t </em>and between the concentrations of
to <em>[A].</em>
![\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E0_t%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%5Cint%5Climits%5EA_%7B0%7D%20_A%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Solving the integral
![\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20k.t%20%2B%20%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D)
Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
<span>1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced</span>
D.) There must be an equal number of atoms of each element on both sides of the equation