Q.30 Derive the relation between half life period and rate constant for first order reaction.
Write the net cell reaction during discharging of lead accumulator.
Draw the structure of peroxymonosulphuric acid.
Best answer
Answer: \(t_{1/2} = \dfrac{0.693}{k}\)
Half-life of a first order reaction: The integrated rate law is \(k = \dfrac{2.303}{t}\log_{10}\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\).
At \(t = t_{1/2}\), \([\mathrm{A}]_t = \dfrac{[\mathrm{A}]_0}{2}\):
\(k = \dfrac{2.303}{t_{1/2}}\log_{10}\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_0/2} = \dfrac{2.303}{t_{1/2}}\log_{10}2 = \dfrac{2.303\times0.3010}{t_{1/2}}\)
\(\therefore t_{1/2} = \dfrac{0.693}{k}\); the half-life of a first order reaction is independent of the initial concentration.
Lead accumulator, net cell reaction on discharge:
Anode: Pb(s) + SO\(_4^{2-}\)(aq) \(\rightarrow\) PbSO\(_4\)(s) + 2e\(^-\)
Cathode: PbO\(_2\)(s) + 4H\(^+\)(aq) + SO\(_4^{2-}\)(aq) + 2e\(^-\) \(\rightarrow\) PbSO\(_4\)(s) + 2H\(_2\)O(l)
Net: Pb(s) + PbO\(_2\)(s) + 2H\(_2\)SO\(_4\)(aq) \(\rightarrow\) 2PbSO\(_4\)(s) + 2H\(_2\)O(l)
Peroxymonosulphuric acid (Caro's acid), H\(_2\)SO\(_5\): [Structure: a central S atom bonded to two O atoms by double bonds (S=O), to one –OH group, and to an –O–O–H (peroxy) group; i.e. HO–S(=O)\(_2\)–O–O–H. The S atom is tetrahedral and contains one peroxide (–O–O–) linkage.]
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