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Ammonium carbamate (NH_2COONH_4)decomposes as: NH_2COONH_4(s) hArr 2NH_3(g) + CO_2(g) In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that partial pressure of NH_3 becomes equal to original total pressure. Calculate the ratio of final total pressure to the original pressure.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/suppose-656311" style="font-weight:bold;" target="_blank" title="Click to know more about SUPPOSE">SUPPOSE</a> the total pressure of the mixture initially is p. This is <a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> to the <a href="https://interviewquestions.tuteehub.com/tag/partial-596556" style="font-weight:bold;" target="_blank" title="Click to know more about PARTIAL">PARTIAL</a> pressure of `NH_3` and `CO_2` which are present in the ratio of 2:1. <br/> `p(NH_3)=2/3p` and `p(CO_2)=1/3p` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> K_p=p(NH_3)^2 xx p(CO_2)=(2/3p)^<a href="https://interviewquestions.tuteehub.com/tag/2xx-1840186" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX">2XX</a>(1/3p)=4/27p^3` <br/> After adding`NH_3 , p(NH_3)=p` (given) <br/> `therefore p^2xxp(CO_2)=K_p=4/27p^3` <br/> `therefore p(CO_2)=4/27p` <br/> Final total pressure =`p(NH_3)+p(CO_2)=p+4/27p =31/27p` <br/> `therefore` Ratio of final total pressure to the original pressure =31:27</body></html>


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