Answer:
The correct answer is 160.37 KJ/mol.
Explanation:
To find the activation energy in the given case, there is a need to use the Arrhenius equation, which is,
k = Ae^-Ea/RT
k1 = Ae^-Ea/RT1 and k2 = Ae^-Ea/RT2
k2/k1 = e^-Ea/R (1/T2-1/T1)
ln(k2/k1) = Ea/R (1/T1-1/T2)
The values of rate constant k1 and k2 are 3.61 * 10^-15 s^-1 and 8.66 * 10^-7 s^-1.
The temperatures T1 and T2 are 298 K and 425 K respectively.
Now by filling the values we get:
ln (8.66*10^-7/3.61*10^-15) = Ea/R (1/298-1/425)
19.29 = Ea/R * 0.001
Ea = 160.37 KJ/mol
You have to use everything that is given since you have to know which is the limiting reactant. We find the limiting reactant by calculating the number of moles of each reactant and compare the number of moles. The limiting reactant would be the one that is consumed fully by the reaction.
Answer:
a). Coordination no. of ![[Ni(en)_2(NH_3)_2]^2+](https://tex.z-dn.net/?f=%5BNi%28en%29_2%28NH_3%29_2%5D%5E2%2B)
= 6
b). Coordination no. of ![[Fe(en)(ox)Cl_2]^2-](https://tex.z-dn.net/?f=%5BFe%28en%29%28ox%29Cl_2%5D%5E2-)
= 6
Explanation:
Coordination number is defined as number of donar atoms bonded to the central atom of the complex ion.
a). Coordination no. of = 6
en or ethylenediamine is a bidentate ligand.
In bidentate ligand, two atoms directly bonded to the central atom.
NH3 is a unidentate ligand.
So, coordination no.= No. of bidentate ligand x 2 + No. of unidentate ligand
= 
b). Coordination no. of = 6
Ethylenediamine (en) is a bidentate ligand.
oxalate ion (ox) is also a bidentate ligand.
Cl is a unidentate ligand
So, coordination no.= No. of bidentate ligand x 2 + No. of unidentate ligand
=
When carbon compounds react with themselves to form a larger molecule the process is called polymerization. Specifically addition polymerization. sometimes molecules containing containing carbon to carbon double bonds can join together to form longer chains. The double bond is broken and the electrons in it join to neighboring molecules.
If the number of valence electron is one than it will be placed in group IA. elements of group IIA will have two valence electrons and so on....
