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  • The half-life of first order decomposition of <img alt="\mathrm{\mathrm{\mathrm{NH}_4 \: \mathrm{NO}_2 \: is \: 2 \mathrm{hr}\: at \: 288 \mathrm{~K}}}" src="/latex-image/?%5Cmathrm%7B%5Cmathrm%7B%

The half-life of first order decomposition of \mathrm{\mathrm{\mathrm{NH}_4 \: \mathrm{NO}_2 \: is \: 2 \mathrm{hr}\: at \: 288 \mathrm{~K}}} temperature \mathrm{\mathrm{NH}_4 \mathrm{NO}_2(\mathrm{aq}) \rightarrow \mathrm{N}_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \: if \, 5.2 \mathrm{~g}\: of\: \mathrm{NH}_4 \mathrm{NO}_2} is allowed to decompose. The time required for \mathrm{NH_4 NO_2 } to decompose \mathrm{80 \% } and the volume of dry \mathrm{\mathrm{N}_2 \mathrm{O} } produced at this point measured at STP are respectively.
 

Option: 1

6.978 \mathrm{hr}, 2.016 \mathrm{~L}
 


Option: 2

4.65 \mathrm{hr}, 1.5 \mathrm{~L}
 


Option: 3

0.319 \mathrm{hr}, 2.016 \mathrm{~L}
 


Option: 4

None of these


Answers (1)

best_answer

\mathrm{ t_{1 / 2}=\frac{0.693}{k} }

\mathrm{ k=\frac{0.693}{2} }

\mathrm{ k=0.3465\: \mathrm{hr}^{-1} }

\mathrm{So, A_0=100 \% }

\mathrm{ A_0-x=20 }

\mathrm{ K=\frac{2.303}{t} \log \left(\frac{A_0}{A_0-x}\right) }

\mathrm{ t=\frac{2.303}{0.3465} \log \left(\frac{100}{20}\right) }

\mathrm{ t=4.65 / \mathrm{hr} }

\mathrm{ n_{\mathrm{NH}_4 \mathrm{NO}_3}=\frac{5.2}{62} }

\mathrm{ n_{\mathrm{NH}_4 \mathrm{NO}_3}=0.0838 }

As per given no. of moles of \mathrm{ \mathrm{N}_2 \mathrm{O} } produced

\mathrm{ =0.0838 \times 80 / 100=0.067 }

Vol of \mathrm{ \mathrm{N}_2 \mathrm{O} } produced at STP

\mathrm{ =22.4 \times 0.067=1.5 L }

Hence option 2 is correct.
 

Posted by

Shailly goel

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