Answer:
The order of reaction is 2.
Rate constant is 0.0328 (M s)⁻¹
Explanation:
The rate of a reaction is inversely proportional to the time taken for the reaction.
As we are decreasing the concentration of the reactant the half life is increasing.
a) For zero order reaction: the half life is directly proportional to initial concentration of reactant
b) for first order reaction: the half life is independent of the initial concentration.
c) higher order reaction: The relation between half life and rate of reaction is:
Rate = ![\frac{1}{k[A_{0}]^{(n-1)}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%5BA_%7B0%7D%5D%5E%7B%28n-1%29%7D%7D)
Half life =![K\frac{1}{[A_{0}]^{(n-1)} }](https://tex.z-dn.net/?f=K%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%5E%7B%28n-1%29%7D%20%7D)
![\frac{(halflife_{1})}{(halflife_{2})}=\frac{[A_{2}]^{(n-1)}}{[A_{1}]^{(n-1)} }](https://tex.z-dn.net/?f=%5Cfrac%7B%28halflife_%7B1%7D%29%7D%7B%28halflife_%7B2%7D%29%7D%3D%5Cfrac%7B%5BA_%7B2%7D%5D%5E%7B%28n-1%29%7D%7D%7B%5BA_%7B1%7D%5D%5E%7B%28n-1%29%7D%20%7D)
where n = order of reaction
Putting values
![\frac{109}{231}=\frac{[0.132]^{(n-1)}}{[0.280]^{(n-1)}}](https://tex.z-dn.net/?f=%5Cfrac%7B109%7D%7B231%7D%3D%5Cfrac%7B%5B0.132%5D%5E%7B%28n-1%29%7D%7D%7B%5B0.280%5D%5E%7B%28n-1%29%7D%7D)
Hence n = 2
![halflife=\frac{1}{k[A_{0}]}](https://tex.z-dn.net/?f=halflife%3D%5Cfrac%7B1%7D%7Bk%5BA_%7B0%7D%5D%7D)
Putting values
K = 0.0328
Moving of positive charge electron and negative charge proton.
A wavelength's frequency and energy (E) drop as it becomes longer. You may conclude from these equations that the wavelength grows shorter as the frequency rises. The wavelength lengthens as the frequency drops.
Since frequency and wavelength are inversely related to one another, the wavelength of the wave falls as frequency increases.
Answer:
The experimenter observed this experiment in a lab rather than natural world because it might be dangerous to the atmosphere if he does the experiment in the natural world and it was still an hypothesis so that's why he did it in the lab.
Pb(NO₃)₂ ⇒limiting reactant
moles PbI₂ = 1.36 x 10⁻³
% yield = 87.72%
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)
Required
- moles of PbI₂
- Limiting reactant
- % yield
Solution
Balanced equation :
Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)
mol Pb(NO₃)₂ :
= 0.45 : 331 g/mol
= 1.36 x 10⁻³
mol NaI :
= 250 ml x 0.25 M
= 0.0625
Limiting reactant (mol : coefficient)
Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³
NaI : 0.0625 : 2 = 0.03125
Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)
moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)
Mass of PbI₂ :
= mol x MW
= 1.36 x 10⁻³ x 461,01 g/mol
= 0.627 g
% yield = 0.55/0.627 x 100% = 87.72%
