Question

$\log 1.5\times 10^7=\log A-\left(\frac{2}{2.303R}\right)$ In the above equation, what is the value of Arrhenius factor?

• A. $3\times 10^6$
• B. $6.3\times 10^9$
• C. $5.42\times 10^{13}$
• D. $5.42\times 10^{10}$

Answer 1

Answer: (D)

$\log k=\log A-\frac{E_a}{2.303RT}$
$\log 1.5\times 10^7=\log A-\frac{22011}{2.303\times 8.314}$
$7+\log 1.5=\log A-3.5591$
$\log A-\log 1.5=10.5591$
$\log \frac{A}{1.5s^{-1}}=10.5591$
$\frac{A}{1.5s^{-1}}=$ anti$-\log (10.5591)=3.615\times 10^{10}$
Now, $A=3.615\times 10^{10}\times 1.5$
$=5.4225\times 10^{10}{s}^{-1}$