Question
Obtain Arrhenius equation, $k = A \times e ^{- E _3 / R T}$
✓
Answer
$(i)$ From experimental observations of variation in rate constants with temperature, Arrhenius developed a mathematical equation between reaction rate constant $( k )$, activation energy $\left( E _{ a }\right)$ and temperature $T$.
$(ii)$
When a graph of Ink is plotted against reciprocal of temperature $(1 / T )$ a straight line with a negative slope is obtained. This is described by a mathematical equation as,
$\ln k=\ln A-\frac{E_{ a }}{R T}$
$=\ln A-\frac{E_{ a }}{R T} \ln e$
$\because \ln e=\log _{ e } e=1$
$=\ln A+\left(\frac{-E_{ a }}{R T}\right) \ln e_{}$
$=\ln A+\ln e ^{-E_{ a } / R T}$
$\therefore \log _{ e } k=\log _{ e } A+\log _{ e } e ^{-E_{ a } / R T}$
$\quad=\log _{ e } A \times e ^{-E_{ a } / R T}$
$\therefore 2.303 \log _{10} k=2.303 \log _{10} A \times e ^{-E_{ a } / R T}$
$\therefore \log _{10} k=\log _{10} A \times e ^{-E_{ a } / R T}$
By taking antilogarithm,
$k=A \times e ^{-E_{ a } / R T}$
$(ii)$
When a graph of Ink is plotted against reciprocal of temperature $(1 / T )$ a straight line with a negative slope is obtained. This is described by a mathematical equation as,
$\ln k=\ln A-\frac{E_{ a }}{R T}$
$=\ln A-\frac{E_{ a }}{R T} \ln e$
$\because \ln e=\log _{ e } e=1$
$=\ln A+\left(\frac{-E_{ a }}{R T}\right) \ln e_{}$
$=\ln A+\ln e ^{-E_{ a } / R T}$
$\therefore \log _{ e } k=\log _{ e } A+\log _{ e } e ^{-E_{ a } / R T}$
$\quad=\log _{ e } A \times e ^{-E_{ a } / R T}$
$\therefore 2.303 \log _{10} k=2.303 \log _{10} A \times e ^{-E_{ a } / R T}$
$\therefore \log _{10} k=\log _{10} A \times e ^{-E_{ a } / R T}$
By taking antilogarithm,
$k=A \times e ^{-E_{ a } / R T}$
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