A arrow B The rate constants of the above reaction at 200 K and 300 K are 0.03 min -1 and 0.05 min -1 respectively. The activation energy for the reaction is J (Nearest integer) (Given: ln 10=2.3 R =8.3 JK -1 mol -1 log 5=0.70 log 3=0.48 log 2=0.30)

Question

A arrow B The rate constants of the above reaction at 200 K and 300 K are 0.03 min -1 and 0.05 min -1 respectively. The activation energy for the reaction is J (Nearest integer) (Given: ln 10=2.3 R =8.3 JK -1 mol -1 log 5=0.70 log 3=0.48 log 2=0.30) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

log ( K 300/ K 200)=( E a /2.3 × 8.314)((1/ T 1)-(1/ T 2)) log (0.05/0.03)=( Ea /2.305 × 8.314) ×[(1/200)-(1/300)] E a =2519.88 J ⇒ E a =2520 J

Q. A→B The rate constants of the above reaction at 200K and 300K are 0.03min−1 and 0.05min−1 respectively. The activation energy for the reaction is __J (Nearest integer) (Given :ln10=2.3 R=8.3JK−1mol−1 log5=0.70 log3=0.48 log2=0.30)

logK200​K300​​=2.3×8.314Ea​​(T1​1​−T2​1​) log0.030.05​=2.305×8.314Ea​×[2001​−3001​] Ea​=2519.88J⇒Ea​=2520J

$lo g \frac{K _{300}}{K _{200}} = \frac{E _{a}}{2.3 \times 8.314} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$
$lo g \frac{0.05}{0.03} = \frac{E a}{2.305 \times 8.314} \times [\frac{1}{200} - \frac{1}{300}]$
$E_{a} = 2519.88 J \Rightarrow E_{a} = 2520 J$