consider the following reaction: 2GeO(g) +W2O3(g) 2GeWO4 Kp = 4.7 times 10^3 del
ID: 927996 • Letter: C
Question
consider the following reaction: 2GeO(g) +W2O3(g) 2GeWO4 Kp = 4.7 times 10^3 delta H = 92 KJ T= 425K When the equilibrium partial pressure of GeO(g) = 0.020 atm and GeWO4(g) = 0.98 atm. what is the equilibrium partial pressure of W2O6(g)? Circle ike term in the parentheses to mate correct statements. All questions refer to the equilibrium in question 7. If the volume is doubled at constant temperature and pressure, the equilibrium will: (shift right, shift left, stay the same) If the temperature is doubled at constant volume and pressure, the equilibrium will. (shift right, shift left, stay the same) 0 If GeO is added at constant temperature and pressure, the equilibrium will: (shift right, shift left, stay the same)Explanation / Answer
7.
(a) Kp = [GeWO4]^2/[GeO]^2[W2O6]
(b) Kc = Kp/(RT)^dn
dn = 2-3 = -1
Kc = 4.7 x 10^3/(0.08205 x 425)^-1 = 1.64 x 10^5
(c) [W2O6] = (0.98)^2/(0.020)^2(4.7 x 10^3) = 0.511 atm
d) If volume is doubled while T and P are same, the equilibrium will : shift right
e) If temperature is doubled while P and V are same, the equilibrium will: shift left
endothermic reaction favor high temperature
f) If GeO is added the equilibrium will: shift left
LeChatellier's principle, as amount increase on one side, it will sift to other side to reequilibriate.
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