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A small amount of NH_(4)HS is placed in a flask already containing ammonia gas at a certain temperature and 0*50 atm pressure. Ammonium hydrogensulphide decomposes to yield NH_(3) andH_(2)S gass in the flask . When the decomposition reaction reaches equilibrium , the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH_(4)HS decomposition at this temperature is

Answer» <html><body><p>`0.30`<br/>`0*18`<br/>`0*<a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>` <br/>`0*<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>`</p>Solution :`{:(,NH_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)HS (s),<a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a>,NH_(3) (g),+,H_(2)S (g)), ("Intial",a " moles",,0.5" atm",,), (" At .eqm.",(a-<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>),,(0.5+p),,p " atm".):}` <br/> i.e., if x moles of `NH_(4)HS` decompose , increase in pressure due to `NH_(3)` = increase in pressure due to `H_(2)S` = p atm .<br/> Total pressure at equilibrium <br/> ` 0.5 + p+ p = 0.5 + 2 p ` atm <br/> ` 0.5 + 2 p = 0.84 " atm" or p=0.17` atm <br/> ` :. p_(NH_(3)) = 0.5 + 0.17 = 0.67` atm<br/> ` p_(H_(2)S) = 0.17 ` atm <br/> ` K_(p) =p_(NH_(3)) xx p_(H_(2)S)` <br/> ` =0.67 xx 0.17 = 0.1139`</body></html>


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