Rate constant k varies with temperature by equation: lo g k ( mi n − 1 ) = 5 − T 2000 K we can conclude:

lo g k = lo g a − 2.303 RT E a 2.303 × R E a = 2000 c a l ∴ E a = 2000 × 2.303 × R × T E a = 2000 × 2.303 × 2 × 1 0 − 3 k c a l E a = 9.212 k c a l

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ChemistryChemical KineticsArrhenius EquationMedium2 minQB

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Rate constant k varies with temperature by equation:

$lo g k (mi n^{- 1}) = 5 - \frac{2000 K}{T}$

we can conclude:

Answer:

Solution:

$lo g k = lo g a - \frac{E _{a}}{2.303 RT}$

$\frac{E _{a}}{2.303 \times R} = 2000 c a l$

∴ $E_{a} = 2000 \times 2.303 \times R \times T$

$E_{a} = 2000 \times 2.303 \times 2 \times 1 0^{- 3} k c a l$

$E_{a} = 9.212 k c a l$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Arrhenius Equation

⏱ 2mℹ️ Source: QB

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