The slope of ln K vs \frac{1}{T} was found to be – 40 if 'T' is in kelvin. The value of ln µ160K will be : (µT represents temperature coefficient at T Kelvin)

Engineering

Chemistry

Arrhenius Equation

Question

The slope of ln K vs $\frac{1}{T}$ was found to be – 40 if 'T' is in kelvin. The value of ln µ_160K will be :
(µ_T represents temperature coefficient at T Kelvin)

$\frac{1}{17}$

$\frac{1}{68}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$ℓnK = ℓnA - \frac{Ea}{RT}$ & $μ = \frac{K_{170}}{K_{160}}$

$slope = \frac{Ea}{R}$

$\frac{- Ea}{R} = - 40$ E_a = 40R

$ℓn (\frac{K_{170}}{K_{160}}) = \frac{Ea}{R} (\frac{1}{160} - \frac{1}{170})$

$ℓn μ = \frac{40 R}{R} (\frac{10}{160 \times 170}) = \frac{1}{4 \times 17} = \frac{1}{68}$