Order Of Reaction

To solve the problem, we need to determine how the rate of a second-order reaction is affected when the concentration of the reactant is (i) doubled and (ii) reduced to half.

1. Understand the Rate Law:
For a second-order reaction with respect to a reactant A, the rate law is , where is the rate constant and is the concentration of the reactant.

2. Case (i) - Concentration Doubled:
If the concentration is doubled, becomes . The new rate is:

.
The rate increases by a factor of 4.

3. Case (ii) - Concentration Reduced to Half:
If the concentration is reduced to half, becomes . The new rate is:

.
The rate decreases to one-fourth of the initial rate.

Final Answer:
(i) The rate increases by a factor of 4 when the concentration is doubled.
(ii) The rate decreases to one-fourth when the concentration is reduced to half.

To solve the problem, we need to write the IUPAC names of the given coordination entities: (a) , (b) , and (c) .

1. Name :
- Central metal: Fe, oxidation state: +3 (since 2 Cl⁻ give -2, total charge +1, so Fe is +3).
- Ligands: en (ethylenediamine, neutral) × 2, Cl⁻ × 2.
- Name ligands alphabetically: ‘en’ (di- for 2) as ethylenediamine, Cl as chloro.
- Complex is cationic, so metal name is iron(III).
IUPAC name: .

2. Name :
- Central metal: Co, oxidation state: +3 (Br⁻ gives -1, gives -2, total charge 0, so Co is +3).
- Ligands in complex: (neutral) × 4, (neutral) × 1, Br⁻ × 1.
- Name ligands alphabetically: ammine ( , tetra- for 4), aqua ( ), bromido (Br⁻).
- Complex is cationic, counter ion is sulfate, metal name is cobalt(III).
IUPAC name: .

3. Name :
- Central metal: Ni, oxidation state: +2 (4 CN⁻ give -4, total charge -2, so Ni is +2).
- Ligands: CN⁻ × 4, named as cyano (but in IUPAC, ‘cyano’ is used, tetra- for 4).
- Complex is anionic, so metal name is nickelate(II).
IUPAC name: .

Final Answer:
(a) , (b) , (c) .