Assume that the decomposition of HNO_(3) can be represented by the following equation 4HNO_(3(g)) hArr 4NO_(2(g)) + 2H_(2) O_((g)) + O_(2(g))and the reaction approaches equilibrium at 400K temperature and the copper turnning 0 atm pressure. At cquilibrium partial pressure of HNO_(3) is 2 atm. Calculate Kc in ("mole" //L)^(3) at 400 K.

Assume that the decomposition of HNO_(3) can be represented by the fol

1.	Assume that the decomposition of HNO_(3) can be represented by the following equation 4HNO_(3(g)) hArr 4NO_(2(g)) + 2H_(2) O_((g)) + O_(2(g))and the reaction approaches equilibrium at 400K temperature and the copper turnning 0 atm pressure. At cquilibrium partial pressure of HNO_(3) is 2 atm. Calculate Kc in ("mole" //L)^(3) at 400 K.
Answer» 4 8 16 32 Solution :`4HNO_(3(g)) harr 4NO_(2(g))+2H_(2)O_((g))+O_(2(g))` `P_("Total")=P_(HNO_(3))+P_(NO_(2))+P_(H_(2)O)+P_(O_(2))` `:. P_(NO_(2))=4_(PO_(2))` and `P_(H_(2)O)=2P_(O_(2))` `:. P_("Total")=P_(HNO_(3))+7P_(O_(2))` `IMPLIES 30-2=P_(O_(2)) xx 7, P_(O_(2))=4` `Kp=(P_(NO_(2))^(4) xx P_(H_(2)O) xx P_(O_(2)))/(P_(HNO_(3))^(4))` `=((4 xx 4) xx (2 xx 4)^(2) xx 4)/(2^(4))=2^(20)` `Kp=Kc(RT)^(Delta n_((g)))=Kc(0.0821 xx 400)^(3)` `implies Kc(0.08 xx 400)^(3) implies Kc=32`