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 N2 (g) + 3H2 (g)   2 NH3 (g) ∆H = - 91 kJ DECREASED PRESSURE, Shift to MOST moles of gas. Most moles on Left: 1 + 3 = 4 moles 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

What can be done to drive the reaction to the LEFT? 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