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Simple Interest

For the react...

Question

For the reaction, $a A + b B \to c C + d D,$ the plot of $log K$ vs $1 / T$ is given below:

The temperature at which the rate constant of the reaction is
$10^{- 4} s^{- 1}$ is ________ $K .$ (Rounded off to the nearest integer)

[Given: The rate constant of the reaction is $10^{- 5} s^{- 1}$ at $500 K .]$

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Solution

From Arrhenius equation,
$l o g_{10} K = l o g_{10} A - \frac{E_{a}}{2.303 R T}$

$Slope = - \frac{E_{a}}{2.303 R} = - 10000$

$l o g_{10} \frac{K_{1}}{K_{2}} = - \frac{E_{a}}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l o g_{10} \frac{10^{- 4}}{10^{- 5}} = - 10000 (\frac{1}{T} - \frac{1}{500})$

$1 = 10000 (\frac{1}{500} - \frac{1}{T})$

$\frac{1}{10000} = \frac{1}{500} - \frac{1}{T}$

$\frac{1}{T} = \frac{1}{500} - \frac{1}{10000}$

$\frac{1}{T} = \frac{20 - 1}{10000} = \frac{19}{10000}$

$T = \frac{10000}{19}$

$∴ T = 526 K$

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Similar questions

l n k versus 1 / T

500 K

10^{- 4} s^{- 1}

600 K

l n_{e} (10) \approx 2.303

1 / T

400

10^{- 5}

s^{- 1}

500

A \to P

(T)

(k)

log k = - (2000) \frac{1}{T} + 6.0

{kJ mol}^{- 1}

R = 8.3 {J mol}^{- 1} K^{- 1}

570 s

\times 10^{- 3} s^{- 1}

{log}_{10} 2 = 0.301, l n 10 = 2.303

a A + b B \to c C + d D

= \frac{d x}{d t} = \frac{- 1}{a} \frac{d [A]}{d t} = \frac{- 1}{b} \frac{d [B]}{d t} = \frac{- 1}{c} \frac{d [C]}{d t} = \frac{- 1}{d} \frac{d [D]}{d t}

3 B r O^{⊖} \to B r O_{3}^{⊖} + 2 B r^{⊖}

80^{\circ} C

- \frac{- d [B r O^{⊖}]}{d t}

0.054 L m o l^{- 1} s^{- 1}

\frac{d [B r O_{3}^{⊖}]}{d t}

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