AP Questions: putting it all together

Equilbrium Constant and EXTERNAL EFFECTS Formation of stalactites and stalagmites CaCO3(s) + H2O(liq) + CO2(g) fg Ca2+(aq) + 2 HCO3-(aq)

The outcome is governed by LE CHATELIER’S PRINCIPLE EQUILIBRIUM Temperature, catalysts, and changes in concentration/ pressure affect equilibria. The outcome is governed by LE CHATELIER’S PRINCIPLE “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

Equilibrium constant, K, and Concentration Concentration changes Δ [ X ] no change in K only the position of equilibrium changes.

Butane-Isobutane Equilibrium

Butane Isobutane At equilibrium K = 2.5 [iso] = 1.25 M [butane] = 0.50 M An additional 1.50 M butane is added. When the system comes to equilibrium again, what are [iso] and [butane]?

Butane e Isobutane Solution Calculate Q immediately after adding more butane and compare with K. Q is LESS THAN K. Therefore, the reaction will shift to the ____________.

Butane e Isobutane Q is less than K, shifts right toward isobutane. Set up ICE table [butane] [isobutane] Initial Change Equilibrium 0.50 + 1.50 1.25 - X + X 2.00 – x 1.25 + x

Butane e Isobutane x = 1.07 M At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane.

Equilibrium Constant and Catalyst Add catalyst: NO change in K A catalyst only affects the RATE it approach equilibrium. Catalytic exhaust system

Pressure and Equilibrium N2O4(g) e 2 NO2(g) Increase P in the system by reducing the volume (constant Temp). e

Increase P (decreased V). Therefore, reaction shifts LEFT N2O4(g) e 2 NO2(g) Increase P (decreased V). In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). Therefore, reaction shifts LEFT P of N2O4 and P of NO2 .

Temperature Effects on Equilibrium Figure 16.6

Temperature Effects on Equilibrium Kc (273 K) = 0.00077 Kc (298 K) = 0.0059 N2O4 (colorless) + heat e 2 NO2(brown) ∆Ho = + 57.2 kJ (endo)

Every T has a unique K Temperature change = change in K Consider the fizz in a soft drink H2CO3(aq) + HEAT e CO2(g) + H2O(l) K = [CO2] / [H2CO3] Increase T. What happens to equilibrium position? To value of K? K increases as T goes up because [CO2] increases and [H2CO3] decreases. Decrease T. Now what? Equilibrium shifts left and K decreases.

NH3 Production N2(g) + 3 H2(g) e 2 NH3(g) + heat K = 3.5 x 108 at 298 K What is the advantage of using the above system to produce ammonia?

Le Chatelier’s Principle Change T - changes K causes change in P or concentrations at equilibrium Use a catalyst: K not changed. Reaction comes more quickly to equilibrium. Add or take away reactant or product: K does not change Reaction adjusts to new equilibrium “position”