It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be kJ mol -1. [nearest integer] (Given ln 10=2.3, R =8.3 JK -1 mol -1, log 2=0.30 )

Question

It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be kJ mol -1. [nearest integer] (Given ln 10=2.3, R =8.3 JK -1 mol -1, log 2=0.30 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

log 10 ( K 2/ K 1)=( E a /2.303 R )((1/300)-(1/309)) 0.3=( E a /2.303 × 8.3)((9/300 × 309)) E a =(0.3 × 2.303 × 8.3 × 300 × 309/9) =59065.04 J E a =59.06 kJ

Q. It has been found that for a chemical reaction with rise in temperature by 9K the rate constant gets doubled. Assuming a reaction to be occurring at 300K, the value of activation energy is found to be ____kJmol−1. [nearest integer] (Given ln10=2.3,R=8.3JK−1mol−1,log2=0.30 )

log10​K1​K2​​=2.303REa​​(3001​−3091​) 0.3=2.303×8.3Ea​​(300×3099​) Ea​=90.3×2.303×8.3×300×309​ =59065.04J Ea​=59.06kJ

$lo g_{10} \frac{K _{2}}{K _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{300} - \frac{1}{309})$
$0.3 = \frac{E _{a}}{2.303 \times 8.3} (\frac{9}{300 \times 309})$
$E_{a} = \frac{0.3 \times 2.303 \times 8.3 \times 300 \times 309}{9}$
$= 59065.04 J$
$E_{a} = 59.06 k J$