Question

Rate constant varies with temperature by the equation $lo{g}_{10}K=5–2000/T$. We can conclude that $(R=8.314Jmo{l}^{-1}{K}^{-1})$:-

• A. Pre exponential factor A is 5
• B. $E_a$ is 4 cal/mol
• C. Pre exponential factor A is ${10}^5$
• D. $E_a$ is 19.212 cal/mol

Answer 1

The correct option is C Pre exponential factor A is ${10}^5$

log K =log A -$\frac{E_a}{2.303RT}$
log A =5
$A={10}^5$