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Ammonia under a pressure of 1.5 atm at 27^(@)C is heated to 374^(@)C in a closed vessel in the presence of a catalyst. Under the conditions, NH_(3) is partially decomposed according to the equation. 2NH_(3) hArr N_(2) + 3H_(2) the vessel is such that the volume remains etfectively constant where as pressure increases to 50 atm. Calculate the percentage of NH_(3) actually decomposed

Answer» <html><body><p>0.065<br/>0.613<br/>0.625<br/>0.64</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E03_042_S01.png" width="80%"/> <br/> Initial pressure of `NH_(3)` of .a. <br/> Mole = 15 atm at `27^(@)C` <br/> The pressure of .a. mole of `NH_(3)`=p atm at = `347^(@)C` <br/> `(P_(1))/(T_(1))=(P_(2))/(T_(2)), (15)/(300)=(P)/(<a href="https://interviewquestions.tuteehub.com/tag/629-1908877" style="font-weight:bold;" target="_blank" title="Click to know more about 629">629</a>) implies p =31` atm <br/> At constant volume and at `347^(@)C` mole `alpha` pressure a `alpha` 31 (before equilibrium) <br/> `(a+2alpha)alpha50` (at equilibrium) `(a+2alpha)/(a)=(50)/(31) implies x=(19a)/(<a href="https://interviewquestions.tuteehub.com/tag/62-330265" style="font-weight:bold;" target="_blank" title="Click to know more about 62">62</a>) implies %NH_(3)` decomposes <br/> `=(2x)/(a) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 1000, (2 xx 19a)/(62 xx a) xx 100=61.3%`</body></html>


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