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STD 12 - 3. Chemical kinetics — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 3. Chemical kinetics4 Marks
MCQ
The temperature dependence of the rate constant $k$ is expressed as $k\, =\, A\,\, e^{-E_a /RT}$ . When a plot between $log\, k$ and $1/T$ is plotted we get the graph as shown. What is the value of slope in the graph ?
  • A
    $\frac{{{E_a}}}{{RT}}$
  • ✓
    $- \frac{{{E_a}}}{{2.303R}}$
  • C
    $ - \frac{{{E_a}}}{{2.303RT}}\log \,A$
  • D
    $- \frac{{{E_a}}}{{2.303}}\,\frac{R}{T}$
✓

Answer

Correct option: B.
$- \frac{{{E_a}}}{{2.303R}}$
b
$K = Ae ^{- E _a / R T}$

$\ln K =\ln \left( Ae ^{- E _{ a } / R T}\right)$

$\ln K =\ln A +\left(-\frac{ E _{ a }}{ RT }\right) \ln e$

$\ln K =\ln A -\frac{ E _{ a }}{ RT }...(1)$

now changing $\ln$ to $\log$

$\log K =\log A +\left(-\frac{ E _{ a }}{2.303 R }\right) \frac{1}{ T }...(2)$

this represent straight time as: $y = c +( m ) x$

$\therefore$ slope $=-\frac{ E _{ a }}{2.303 R }$

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