Q.(A) The rate of a first order reaction, A→B is 5.4 × 10⁻⁶ Ms⁻¹ when [A] is 0.3 M. Calculate the rate constant of the reaction.

A first-order rate-constant calculation, and the periodic trends across O, S, Se, Te, Po in Group 16. OR: a half-life calculation and the classification of oxides.

(A) Rate constant: for a first order reaction, rate $= k[A]$.

$5.4\times10^{-6} = k \times 0.3 \Rightarrow k = \dfrac{5.4\times10^{-6}}{0.3} = 1.8\times10^{-5}\ s^{-1}$

(B) Group 16 properties down the group (O → S → Se → Te → Po):

1. Electronegativity: decreases down the group (oxygen is the most electronegative, second only to fluorine overall), due to increasing atomic size and shielding.
2. Melting and boiling points: generally increase from O to Te (Po is anomalously lower/less regular); oxygen, being a small diatomic molecule ($O_2$) held only by weak van der Waals forces, has a very low mp/bp compared to S, Se, Te, which exist as larger polyatomic/network structures ($S_8$ rings, etc.) with stronger interparticle forces.
3. Metallic character: increases down the group — O, S, Se are non-metals, Te is a metalloid, and Po is essentially metallic — as atomic size increases and ionisation enthalpy decreases.
4. Allotropy: every group 16 element shows allotropy — e.g. $O_2$/$O_3$ for oxygen; rhombic and monoclinic sulphur; red and grey selenium; crystalline and amorphous forms of tellurium. OR alternative: (A) $t_{1/2} = \dfrac{0.693}{k} \Rightarrow k = \dfrac{0.693}{6.0\ h} = 0.1155\ h^{-1}$ ($\approx 3.21\times10^{-5}\ s^{-1}$) (B) Oxides are binary compounds of oxygen with another element, classified as: …