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The half time of first order decompositi...

The half time of first order decomposition of nitramide is `2.1` hour at `15^(@)C`.
`NH_(2)NO_(2(aq.))rarr N_(2)O_((g))+H_(2)O_((l))`
If `6.2 g` of `NH_(2)NO_(2)` is allowed to decompose, calculate:
(i) Time taken for `NH_(2)NO_(2)` is decompose `99%`.
(ii) Volume of dry `N_(2)O` produced at this point measured at STP.

Text Solution

Verified by Experts

`(i) k=(0693)/(t_(1//2))=(0.693)/(2.1)=0.33hr^(-1)`
Applying kinetic eqution of rist order reaction,
`k=(2.303)/(t)log_(10).(a)/((a-x))`
`or t=(2.303)/(0.33)log_(10).(100)/((10099))=13.96hrs`
(ii) No. of moles of `NH_(2)NO_(2)` decomposed
`=0.99xx(6.2)/(62)=0.099`
No. of moels of `N_(2)O` formed =0.099
Volume f `N_(2)O` at STP `=0.099xx225400 mL=2217.6mL`
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