(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
$(R = 8.314\, J \,mol^{-1}\, K^{-1})$
Expt. No. | $(A)$ | $(B)$ | પ્રારંભિક દર |
$1$ | $0.012$ | $0.035$ | $0.10$ |
$2$ | $0.024$ | $0.070$ | $0.80$ |
$3$ |
$0.024$ |
$0.035$ | $0.10$ |
$4$ | $0.012$ | $0.070$ | $0.80$ |