Consider the following reaction where Kc = 2.90x10-at 1150 K. 2SO3(g)2so2(g) + O

Question

Consider the following reaction where Kc = 2.90x10-at 1150 K. 2SO3(g)2so2(g) + O2(g) A reaction mixture was found to contain 4.72x10- moles of S03(g) 2.95x10 moles of SO2(g), and 4.32x10*moles of 02(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. Submit Answer

Explanation / Answer

1. Q= [SO2]2[O2]/ [SO3]2 = (2.95x10*-2)2 (4.32x10*-2)/ (4.72x10*-2)2

Qc= 7.96x10*-2 > Kc (2.90x10*-2)

Since Qc > Kc, since reaction moves in reverse direction ie., forward reaction left (reactant) side.

2. Q=[SO2]2[O2] /[SO3]2

Q= 12 *1/12 = 1 (since pressure of each gas is 1atm)

Q< Kp(2.74)

1. FALSE, since Q<Kp

2. False since it increases

3. True since reaction moves in forward direction hence concentration of so2 increases

4. False

5. False, since reaction moves in forward direction

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