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For a reaction, consider the plot of In ...

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 :

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The correct Answer is:
4
ln k `= ln A - (E_(a))/(RT), y = mx + c`
`|"Slope"| = +4606 xx ((500 - 400))/(500 xx 400)`
`ln((k_(2))/(10^(-5))) = 2.303, ln((k_(2))/(10^(-5))) = ln10`
`k = 10^(-4)`
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