Consider the endothermic reaction: 4 HCI(g) + SiO_2(s) SiCI_4(g) + 2 H_2O(l) At
ID: 937585 • Letter: C
Question
Consider the endothermic reaction: 4 HCI(g) + SiO_2(s) SiCI_4(g) + 2 H_2O(l) At equilibrium, 4.50 mol SiCI_4, 9.00 mol H.O, 2.25 mol 50.00 L container at 25 degree C. Calculate the value for K_c at 25 degree C for the reaction shown above, Calculate the value for Kp at 25 degree C for the reaction shown above. If the phase for water were gaseous, instead of the liquid phase shown above, would the value for K_p at 25 C for the reaction above then be equal to, different from, or cannot be determined when compared with the value for K_c that you calculated in Part a above? Be complete in justifying your choice. If the system described by the equation above is initially at equilibrium, predict the direction of the shift in the position of the equilibrium (left, right, no effect, or cannot be determined) and justify your prediction for each of the following. Some gaseous HCI is removed; temperature and container volume remain constant. Some gaseous SICI_4 is added; temperature and container volume remain constant. To the container is added 2.0 cm of solid SI0_4; temperature and container volume remain constant. From the container is removed 0.50 mol of liquid water; temperature and container volume remain constant. To the container is added 0.20 g of solid platinum to act as a catalyst; temperature and container volume remain constant. Container volume is doubled; temperature remains constant. vii. Temperature is lowered; container volume remains constant. Container volume is halved and temperature is lowered. To the container is added 1.00 mol of gaseous helium; temperature and container volume remain constant. For each of the changes shown in Part d above, what effect (decrease, increase, no effect, or cannot be determined) would there be on the value for the equilibrium constant for the reaction?Explanation / Answer
[HCl] = 4.50 mol / 50 L = 0.09 M
Kc = [SiCl4] / [HCl]4
Kc = 0.09 M / (0.09 M)4
Kc = 1371.74
PSiCl4 = [SiCl4] x R x T
PSiCl4 = 0.09 mol/L x 0.082 L.atm/K.mol x 298.15K
PSiCl4 = 2.200 atm
Kp = P[SiCl4] / P[HCl]4
Kp = 2.200 atm / (2.200 atm)4
Kp = 0.0939
PH2O = 0.18 mol /L x 0.082 L.atm/K.mol x 298.15K
Kp = P[SiCl4] x P[H2O]2 / P[HCl]4
Kp = 2.200 atm x (4.400 atm)2 / (2.200 atm)4
Kp = 1.8182
Kp = Kc x (R x T)n
Kc = 1.8182 / (0.082 atm.L/K.mol x 298.15K)-1
Kc = 44.45
By Le Chatelier’s principle when you add more products, then equilibrium is displaced to the reagents (left) and Kc value is less.
II) Equilibrium will be displaced to the left (reagents) by Le Chatelier’s principle
III) Equilibrium will be displaced to the right (products) by Le Chatelier’s principle
IV) Equilibrium will be displaced to the right (products) by Le Chatelier’s principle
V) No effect, because catalyst doesn’t act in the equilibrium constant, it only increases the reaction rate
VI) Equilibrium will be displaced to the left (reagents) by Le Chatelier’s principle
VII) In theory, equilibrium will be displaced to the right (products) by Le Chatelier’s principle, but it’s important to know if the reaction is endothermic or exothermic, because heat should be considered like a reagent or a product respectively.
VIII) In theory, equilibrium will be displaced to the right (products).
IX) In theory, equilibrium will be displaced to the right (products).
Note: Le Chatelier’s Principle “When any system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established”