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))

A sample of HI_((g))is placed in flask at a pressure of 0.2 atm. At eq

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» SOLUTION :`{:("Reaction:", 2HI_((g)) hArr, H_(2(g)) + , I_(2(g))),("Initial PRESSURE:","0.2 atm",zero,zero),("Change in reaction :", "-2x atm","x atm","x atm"),("Partial pressure at EQUILIBRIUM :", (0.2-2x), "x atm","x atm"),(,=0.04 "atm", =0.08 "atm" , =0.08 "atm"):}` At equilibrium partial pressure of HI = 0.2 - 2x = 0.04 `therefore` 2x=0.2 - 0.04 =0.16 `therefore` x=0.08 atm Thus, `p_(HI)`= 0.04 , `p_(H_2)` = 0.08 = `p_(I_2)`atm `K_p=((p_(H_2))(p_(I_2)))/((p_(HI))^2)="(0.08 atm)(0.08 atm)"/"(0.04 atm)"^2=4.0`