Question

Velocity constant of a reaction at $290\, K$ was found to be $3.2 \times 10^{-3}$. At $300\, K$ it will be

• A. $1.28 \times 10^{-2}$
• B. $9.6 \times 10^{-3}$
• C. $6.4 \times 10^{-3}$
• D. $3.2 \times 10^{-4}$

Answer 1

Answer: (C)

As we know that the velocity constant become double by increasing the temperature by $10^{\circ} C$ so if at $290\, K$.
Velocity constant $=3.2 \times 10^{-3}$ then at $300\, K$ velocity constant
$=2\left( K _{290}\right)=2 \times 3.2 \times$
$10^{-3}=6.4 \times 10^{-3}$.