The reaction is 2 NO (g) <----> N2(g) + O2
partial pressures
Initial 37.30 0 0
Change -2p +p +p
Equilibrium 37.30-p p p
Kp = pN2 X pO2 / (pNO)^2
2400 = p^2 / (37.30-p)^2
3339096 - 179040p + 2400p^2 = p^2
2399p^2 + 3339096 -179040 p = 0
On solving
p = 36.55atm
Thus partial pressure of N2 and O2 = 36. 55 atm
Inertia is the answer to this question
Answer:
The heat gain by the system,
q
=
−
250
kJ
.
The work done on the system ,
w
=
−
500
kJ
.
The First Law of Thermodynamics state that
Δ
U
=
q
+
w
=
−
750
kJ
Explanation:
The balanced equation for the above reaction is;
4NH₃ + 7O₂ --> 4NO₂ + 6H₂O
stoichiometry of NH₃ to NO₂ is 4:4
the number of NH₃ moles consumed are - 4.0 g / 17 g/mol = 0.24 mol
number of NH₃ moles reacted are equivalent to number of NO₂ moles formed
therefore number of NO₂ moles formed - 0.24 mol x 46 g/mol = 11.04 g
mass of NO₂ formed is 11.04 g
