The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700-1000 K. The data has been analysed by plotting In k vs (103/ T ) graph. The value of activation energy for the reaction is kJ mol -1. (Nearest integer) (Given: R =8.31 J K -1 mol -1 )

Question

The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700-1000 K. The data has been analysed by plotting In k vs (103/ T ) graph. The value of activation energy for the reaction is kJ mol -1. (Nearest integer) (Given: R =8.31 J K -1 mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

ln k = ln A -( Ea /103 RT ) × 103 = nA +(103/ T )[-( Ea /103 RT )] From the graph (- Ea /103 × R )=-18.5 Ea =153.735 kJ / mol ∼ 154

Q. The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700−1000K. The data has been analysed by plotting In k vs T103​ graph. The value of activation energy for the reaction is ____ kJmol−1. (Nearest integer) (Given : R=8.31JK−1mol−1 )

lnk=lnA−103RTEa​×103 =nA+T103​[−103RTEa​] From the graph 103×R−Ea​=−18.5 Ea=153.735kJ/mol ∼154

$ln k = ln A - \frac{E a}{1 0 ^{3} RT} \times 1 0^{3}$
$= n A + \frac{1 0 ^{3}}{T} [- \frac{E a}{1 0 ^{3} RT}]$
From the graph
$\frac{- E a}{1 0 ^{3} \times R} = - 18.5$
$E a = 153.735 k J / m o l$
$\sim 154$