(i) Half life,
= \(\frac { 0.693 }{ 200{ s }^{ -1 } } \)=  3.47s (approximately)
(ii) Half life
= \(\frac { 0.693 }{ 2{ min }^{ -1 } } \) = 0.35 min (approximately)
(iii) Half life,
= \(\frac { 0.693 }{ 2{ 4years }^{ -1 } } \)= 0.173 years (approximately)

(i) The rate constant is nearly doubled with a rise in temperature by 10^oC for a chemical reaction.
(ii) The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation.
\(k=A{ e }^{ \frac { -{ E }_{ a } }{ RT } }\)

\(2{ NH }_{ { 3 }_{ (g) } }\overset { pt }{ \longrightarrow } { N }_{ { 2 }_{ (g) } }+3{ H }_{ { 2 }_{ (g) } }\)
Rate = \(-\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { N }_{ 2 } \right] }{ dt } \)
However, it is given that the reaction is of zero order. Therefore
\(-\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { N }_{ 2 } \right] }{ dt } k\)
= 2.5 x 10^-4 mol L^-1 S^-1
The rate of production of N₂ is
\(\frac { d\left[ { N }_{ 2 } \right] }{ dt } \)=  2.5 x 10^-4 mol L^-1 S^-1
The rate of production of H₂ is
\(\frac { d\left[ { N }_{ 2 } \right] }{ dt } \)= 3 x 2.5 x 10^-4 mol L^-1 s^-1
= 7.5 x 10^-4 mol L^-1 s^-1.

In the given case
E_a = 209.5 KJ mol^-1 = 209500 J mol^-1
T = 581 K
R = 8.314 JK^-1 mol^-1
Now the fraction of molecules of reactants having energy equal to or greater than activation energy is given as.
x = e^-E_a/RT
\(\Rightarrow \log { x } =-\frac { { E }_{ a } }{ RT } \)
\(\Rightarrow \log { x } =-\frac { { E }_{ a } }{ 2.303RT } \)
\(\Rightarrow \log { x } =\frac { 209500J{ mol }^{ -1 } }{ 2.303\times 8.314J{ K }^{ -1 }{ mol }^{ -1 }\times 581 } \)
= 18.8323
x = Antilogs (18.8323)
= Antilogs \(\overset { \_ \_ }{ 19 } .1677\)
= 1.471 x 10^-19