Consider the following reaction where Kc = 1.29×10-2 at 600 K. COCl2(g) CO(g) +
ID: 1066947 • Letter: C
Question
Consider the following reaction where Kc = 1.29×10-2 at 600 K. COCl2(g) CO(g) + Cl2(g) A reaction mixture was found to contain 0.125 moles of COCl2(g), 2.66×10-2 moles of CO(g), and 4.48×10-2 moles of Cl2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium?
The reaction quotient, Qc, equals .
The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.
2.Consider the following reaction where Kc = 77.5 at 600 K. CO(g) + Cl2(g) COCl2(g) A reaction mixture was found to contain 2.51×10-2 moles of CO(g), 4.29×10-2 moles of Cl2(g) and 0.129 moles of COCl2(g), in a 1.00 liter container.
Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium?
The reaction quotient, Qc, equals .
The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.
Explanation / Answer
KC = 1.29*10^-2
Q = [CO][Cl2] / [COCl2]
Q = (2.66*10^-2)(4.48*10^-2) /(0.125) = 9.533*10^-3
Then
Q < K, therefore this is not in equilibrium, it will form more reactants
Q2.
Q = [Cocl2] / [CO][Cl2]
Q = (0.129)/((2.51*10^-2)(4.29*10^-2)) = 119.8005
since
Q > K, this will favour more production of "products"