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Q.
$\log 1.5 \times 10^{7}=\log A -\left(\frac{2}{2.303 R }\right)$ In the above equation, what is the value of Arrhenius factor?
Chemical Kinetics
Solution:
$\log k=\log A -\frac{ E _{ a }}{2.303 RT }$
$\log 1.5 \times 10^{7}=\log A -\frac{22011}{2.303 \times 8.314}$
$7+\log 1.5=\log A -3.5591$
$\log A -\log 1.5=10.5591$
$\log \frac{ A }{1.5 s ^{-1}}=10.5591$
$\frac{ A }{1.5 s ^{-1}}=$ anti$-\log (10.5591)=3.615 \times 10^{10}$
Now, $A =3.615 \times 10^{10} \times 1.5$
$=5.4225 \times 10^{10} s ^{-1}$