1.

A sample of HI_((g))is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI_((g))is 0.04 atm. What is K_p for thegiven equilibrium ? 2HI_((g)) hArr H_(2(g)) + I_(2(g))

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`{:("Reaction:", 2HI_((g)) hArr, H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>(g)) + , I_(2(g))),("Initial <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a>:","0.2 atm",zero,zero),("Change in reaction :", "-2x atm","x atm","x atm"),("Partial pressure at <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> :", (0.2-2x), "x atm","x atm"),(,=0.04 "atm", =0.08 "atm" , =0.08 "atm"):}` <br/> At equilibrium partial pressure of <a href="https://interviewquestions.tuteehub.com/tag/hi-479908" style="font-weight:bold;" target="_blank" title="Click to know more about HI">HI</a> = 0.2 - 2x = 0.04<br/> `therefore` 2x=0.2 - 0.04 =0.16 <br/> `therefore` x=0.08 atm <br/> Thus, `p_(HI)`= 0.04 , `p_(H_2)` = 0.08 = `p_(I_2)`atm <br/> `K_p=((p_(H_2))(p_(I_2)))/((p_(HI))^2)="(0.08 atm)(0.08 atm)"/"(0.04 atm)"^2=4.0`</body></html>


Discussion

No Comment Found

Related InterviewSolutions