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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=(0.693)/(t_(1//2))=(0.693)/(2.1" hr")=0.33" hr"^(-1)`
`x=99%" of "a=0.99" a"`
`t=(2.303)/(k)log""(a)/(a-x)=(2.303)/(0.33" hr"^(-1))log""(a)/(a-0.99a)=(2.303)/(0.33)log10^(2)=13.96" hours"`
(ii) Amount decomposed `=99%" of "6.2" g "=(99)/(100)xx6.2" g "=6.138" g"`
1 mol `NH_(2)NO_(2)(62" g")" produce "N_(2)O" at STP"=22.4" L"`
6.138 g will produce `N_(2)O` at STP `=(22.4)/(62)xx6.138" L"=2.2176" L"`
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The half-time of the first-order decomposition of nitramide is 2.1 hour at 15^@C , NH_2NO_2(aq) to N_2O(g) + H_2O(l) If 6.2 g of NH_2NO_2 is allowed to decompose, calculate (i) the time taken for 99% decomposition, and (ii) the volume of dry N_2O produced at this point measured at STP.

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Knowledge Check

  • The half-time of the following first order decomposition of nitramide is 2.1 h at 15^(@)C : NH_(2)NO_(2)(aq)rarrN_(2)O(g)+H_(2)O(l) If 6.2 g of nitramide is allowed to decompose then time taken for it to decompose 99%, will be

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