Chemical Kinetics — Chemistry STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryChemical Kinetics3 Marks
Question
Write Arrhenius equation. Derive an expression for temperature variations.
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Answer
Arrhenius equation, $K=A \cdot e^{-E a / R T}$ for temperature variation
$\log _{10} k=\log _{10} A-\frac{E_a}{2.303 R T}$
at two different temperatures $T_1$ and $T_2$ is written as
$\log _{10} k_1=\log _{10} A-\frac{E_a}{2.303 R T_1} \ldots(1)$
and $\log k_2=\log _{10} A -\frac{ E _{ a }}{2.303 RT _2}..(2)$
where $k1$ is the rate constant at $T1$ and $k2$ at $T2.$ The subtraction of equation $(1)$ from equation $(2)$
$`"\log"_{10}"k"_2 - "\log"_{10}"k"_1 = - "E"_"a"/(2.303"RT"_2) + "E"_"a"/(2.303"RT"_1)$
Or $\log _{10} \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$=\frac{ E _{ a }}{2.303 R }\left[\frac{ T _2- T _1}{ T _1 T _2}\right]$
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