Transcript 278-EQUILIBRIUM_II

EQUILIBRIUM II
2/26/07
Le Chatelier Rule for GAS reactions; As the pressure
INCREEASES the equilibrium will shift to the side with the
LEAST number of moles. LOGIC: increased pressure reduces
the volume, the least number of moles of particles has the
least volume .
EXAMPLE: for the reaction
Most moles on Left:
1 + 3 = 4 moles
DECREASED PRESSURE,
Shift to MOST moles of gas.
N2 (g) + 3H2 (g)   2 NH3 (g) ∆H = - 91 kJ
INCREASED PRESSURE Shift
to side with LEAST moles.
Least moles on Right:
2 = 2 moles
EQUILIBRIUM II-U DO IT IN GROUPS NOW !
2/26/07
EXAMPLE: for the reaction
N2 (g) + 3H2 (g)   2 NH3 (g) ∆H = - 91 kJ
What can be done to drive the reaction to the
RIGHT?
COOL the reaction (removes heat from right), REMOVE ammonia
(EQ shifts right to replace), Increase pressure: all shift to right.
N2 (g) + 3H2 (g)   2 NH3 (g) ∆H = - 91 kJ HEAT
EXOTHERMIC,
HEAT ON
RIGHT
EQUILIBRIUM II-U DO IT IN GROUPS NOW
2/26/07
EXAMPLE: for the reaction
N2 (g) + 3H2 (g)   2 NH3 (g) ∆H = - 91 kJ
What can be done to drive the reaction to the LEFT?
Heating or adding NH3 adds to right and shifts left.
Remove N2 or H2 shifts left. Decrease Pressure
drives to side of most moles, LEFT in this reaction.
EXOTHERMIC
:
HEAT ON
RIGHT
N2(g) + 3H2(g)   2 NH3(g) ∆H = - 91 kJ HEAT