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The reaction , 2N(2)O(5)rarr 4NO(2)+O(2)...

The reaction , `2N_(2)O_(5)rarr 4NO_(2)+O_(2)` is first order and takes `24` minutes for `75%` decomposition of `N_(2)O_(5)` at the end of `1` hour after the start of the reaction find the amount of oxide is left.

Text Solution

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The correct Answer is:
3
`K=2.303/t"log"a/(a-x)=2.303/24"log"100/(100-75)`
`=5.77xx10^(-2) min`
Now `t_(1//2)=0.693/K=0.693/(5.77xx10^(-2))=12 min`
No. of half-lives `(n)=60/12=5 ( :' 1 hr =60 min)`
`:.` Amount left=`(1/2)^(n)=(1/2)^(5)=1/32`
`=1/32xx100~~3%`
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