(લો : $\ln 5=1.6094;\left.R =8.314\, J mol ^{-1} K ^{-1}\right)$
As per question $K _{ T _{2}}=5 K _{ T _{1}}$ as molecules activated are increased five times so k will increases $5$ times
Now
$\ln \left(\frac{ K _{ T _{2}}}{ K _{ T _{1}}}\right)=\frac{ Ea }{ R }\left(\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right)$
$\ln 5=\frac{ Ea }{ R }\left(\frac{15}{300 \times 315}\right)$
So $\quad Ea =\frac{1.6094 \times 8.314 \times 300 \times 315}{15}$
$Ea =84297.47$ Joules/mole

| $p ( mm Hg )$ | $50$ | $100$ | $200$ | $400$ |
| સાપેક્ષ $t _{1 / 2}( s )$ | $4$ | $2$ | $1$ | $0.5$ |
પ્રક્રિયાનો ક્રમ શોધો.
| Run | $[A]/mol\,L^{-1}$ | $[B]/mol\,L^{-1}$ | $D$ ઉત્પન્ન થવાનો શરૂઆતનો દર $mol\,L^{-1}\,min^{-1}$ |
| $I.$ | $0.1$ | $0.1$ | $6.0 \times 10^{-3}$ |
| $II.$ | $0.3$ | $0.2$ | $7.2 \times 10^{-2}$ |
| $III.$ | $0.3$ | $0.4$ | $2.88 \times 10^{-1}$ |
| $IV.$ | $0.4$ | $0.1$ | $2.40 \times 10^{-2}$ |
ઉપરની વિગત પરથી નીચેનામાંથી ક્યું સાચુ છે ?