Answer:
Explanation:
GIven that:
The activation energy = 250 kJ
k₁ = 0.380 /M
k₂ = ???
Initial temperature [tex]T_1 =[/tex] 1001 K
Final temperature [tex]T_2 =[/tex] 298 K
Applying the equation of Arrhenius theory.
[tex]In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})[/tex]
where ;
R gas constant = 8.314 J/K/mol
[tex]In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})[/tex]
[tex]In \dfrac{k_2}{0.380 }= -70.8655[/tex]
[tex]\dfrac{k_2}{0.380 }= e^{-70.8655}[/tex]
[tex]\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}[/tex]
[tex]{k_2}= 1.67303256 \times 10^{-31} \times {0.380 }[/tex]
[tex]{k_2}= 6.3575 \times 10^{-32}[/tex] /M .sec
Half life:
At 1001 K.
[tex]t_{1/2} = \dfrac{In_2}{k_1}[/tex]
[tex]t_{1/2} = \dfrac{0.693}{0.38}[/tex]
[tex]t_{1/2} =[/tex] 1.82368 secc
At 298 K:
[tex]t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}[/tex]
[tex]t_{1/2} =1.09 \times 10^{31} \ sec[/tex]
