The variation of rate constant with temperature follows the equation In =30−40T+20T2−30 ln Tbased on this mformation, identify the most appropriate option.

E A at 300 K is approximately equal to 1 . 08 × 10 9 R .

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The variation of rate constant with temperature follows the equation In $= 30 - \frac{40}{T} + 20 T^{2} - 30 \ln T$
based on this mformation, identify the most appropriate option.

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$E_{A}$ will remain constant at all temperatures.

There is some temperature greater than $50 K$ where $E_{A}$ will become negative.

$E_{A}$ at $300 K$ is approximately equal to $1.08 \times 10^{9} R .$

Rate constant will follows Arrhenius equation.

answer is A.

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$\ln k = 30 - \frac{40}{T} + 20 T^{2} - 30 \ln T$

$\frac{d (\ln k)}{d T} = + \frac{40}{T^{2}} + 40 T - \frac{30}{T}$

$= \frac{E_{a}}{R T^{2}}$

$E_{a} = R [40 + 40 T^{3} - 30 T]$

At $T = 300 K, E_{a} ≃ 1.08 \times 10^{9} R$

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