Consider the following reaction where K c = 2.90×10 -2 at 1150 K. 2 SO3(g)--2SO2
ID: 1024207 • Letter: C
Question
Consider the following reaction where Kc = 2.90×10-2 at 1150 K.
2 SO3(g)--2SO2(g)-02(g) moles of O2g), in a A reaction mixture was found to contain 3.92x10moles of SO3(g), 2.37x10* moles of SO2(g). and 4.48x10 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to each 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
For the given reaction,
Kc = [SO2]^2.[O2]/[SO3]^2
with volume = 1 L, mols = M
So,
Qc = (2.37 x 10^-2)^2.(4.48 x 10^-2)/(3.92 x 10^-2)^-2 = 1.64 x 10^-2
the calculated Qc is greater than given Kc of 2.90 x 10^-2, thus,
The reaction is not at equilibrium.
The direction of reaction is towards right handside, that is towards the product end.
the reaction quotient Qc = (2.37 x 10^-2)^2.(4.48 x 10^-2)/(3.92 x 10^-2)^-2 = 1.64 x 10^-2
The reaction,
B. must run in the reverse direction to reach equilibrium.