Answer: see figure attached and explanation below.
Explanation:
1) Chemical equation (given):
Fe + CuCl₂ → Cu + FeCl₂
2) ΔHf reactants: -256 kJ/mol (given)
3) ΔHf products: - 321 kJ/mol (given)
4) ΔH reaction = ΔHf products - ΔHf reactants = - 321 kJ/mol - (- 256 kJ/mol) = - 65 kJ/mol
5) Conclusion:
i) Since ΔHf of products is less (more negative) than ΔHf of reactants, the reaction is exhotermic: the reaction released energy, which is the reason why the products content less potential energy than the reactants.
ii) Then, the energy diagram is the typical one of an exothermic reaction: the products start a certain potential energy level, the energy incrases until reaching the activation energy (the energy barrier to form the activated complex) and then energy decreases until a level below the energy of the reactants.
iii) See the attached figure with such kind of diagram showing the products at a lower level than the reactans
The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,
A(t) = (A(o))(0.5)^(t/h)
where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,
0.89 = (0.5)(t / 5730 years)
The value of t from the equation is 963.34 years.
<em>Answer: 963 years</em>
Answer : The concentration of NOBr after 95 s is, 0.013 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant =
t = time taken = 95 s
[A] = concentration of substance after time 't' = ?
= Initial concentration = 0.86 M
Now put all the given values in above equation, we get:
![0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)](https://tex.z-dn.net/?f=0.80%3D%5Cfrac%7B1%7D%7B95%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.86%29%7D%5Cright%29)
[A] = 0.013 M
Hence, the concentration of NOBr after 95 s is, 0.013 M