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For a reaction, consider the plot of In k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10^-5 s^-1 , then the rate constant at 500 K is :

Option 1)
$10^{-6}\: s^{-1}$
Option 2)
$2 \times 10^{-4}\: s^{-1}$
Option 3)
$10^{-4}\: s^{-1}$
Option 4)
$4 \times10^{-4}\: s^{-1}$

Answers (1)

best_answer

Chemical Kinetics -

The branch of chemistry which predicts the rate and mechanism of a process

-

Effect of Temperature and Activation Energy on rate constant -

$E_{a}= 0$ $k\neq f(T)$

$E_{a}< 0$ $k$ decreases with the increase in T

$E_{a}> 0$ $k$ increases with increase in T

$E_{a}$ = Activation Energy

T = Temperature

$k$ = Rate constant

- wherein

Rate of reaction -

The rate of the reaction during the courses of a reaction in any instant of time is the change in concentration of reacting species. Rate is a positive quantity.

Unit = $mol \:L^{-} \:sec^{-}=\:[concentration][time]^{-}$

- wherein

$R\rightarrow P$

t = t₁

$R_1\rightarrow P_1$

t = t₂

$R_2\rightarrow P_2$

$\Delta t=t_2-t_1 \:\Delta R=[R_2]-[R_1]$

$\DeltaP=[P_2]-[P_1]$

As we have learned

$l_{n}K=l_{n}A-\frac{E_{a}}{R_{T}}=l_{n}A-\frac{4606}{T}$

$ln(\frac{k}{10^{-5}})=(\frac{Ea}{R})\times \frac{500-400}{500\times 400}$

$l_{n}\left ( \frac{K}{10^{-5}} \right )=4606\times \frac{1}{2000}=2.303$

$l_{n}\left ( \frac{K}{10^{-5}} \right )l_{n}10$

$K=10^{-4}$

Option 1)

$10^{-6}\: s^{-1}$

Option 2)

$2 \times 10^{-4}\: s^{-1}$

Option 3)
$10^{-4}\: s^{-1}$
Option 4)

$4 \times10^{-4}\: s^{-1}$

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