Q. The velocity constant of a reaction at 290 K was found to be $3.2 \times 10^{- 3}s^{- 1}$. When the temperature is raised to 310 K, it will be about
A) $6.4 \times 10^{- 3}$
B) $3.2 \times 10^{- 4}$
C) $9.6 \times 10^{- 3}$
D) $1.28 \times 10^{- 2}$
Explanation
(d)
Rate constant almost gets doubled by the increase of $10℃$ in temperature. Hence, the rate constant at 310 K will be
$= 3.2 \times 10^{- 3} \times (2)^{2}$
($\ \because$increase in temperature=20 K)
$= 1.28 \times 10^{- 2}J^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.