Consider the following reaction where Kc = 9.52×10-2 at 350 K. CH4(g) + CCl4(g)
ID: 953482 • Letter: C
Question
Consider the following reaction where Kc = 9.52×10-2 at 350 K. CH4(g) + CCl4(g) 2CH2Cl2(g)
A reaction mixture was found to contain 2.06×10-2 moles of CH4(g), 4.08×10-2 moles of CCl4(g) and 1.11×10-2 moles of CH2Cl2(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
K = CH4*CCl4 / (CH2Cl2)^2 = 9.52*10^-2
Q = CH4*CCl4 / (CH2Cl2)^2 = (2.06*10^-2)(4.08*10^-2)/(1.11*10^-2)^2 = 6.8215
Q = 6.82
NO, since Q is not equal to K, this is not in equilbirium
if
P = 1=1=1
then
K = CH4*CCl4 / (CH2Cl2)^2
Kp = (9.52*10^-2)*(0.082*350)^0
Kp = Kc
Qp = 1*1/1 = 1
since Qp > Kp, the equilibirum shift to products
FALSE 1. A reaction will occur in which CH2Cl2(g) is consumed.this favours products
FALSE 2. Kp will decrease.Kp is a constant
TRUE 3. A reaction will occur in which CH4 is consumed. since products are favoured
TRUE 4. Qp is greater than Kp.
FALSE 5. The reaction is at equilibrium. No further reaction will occur. since Q> K