In the presence of a catalyst, sulfur dioxide reacts with oxygen to form sulfur

Question

In the presence of a catalyst, sulfur dioxide reacts with oxygen to form sulfur trioxide: 2SO_2 (g) + O_2 (g) doubleheadarrow 2SO_3 (g) When 2.00 mol of O_2 and 2.00 mol of SO_2 are placed in a 1 L container and comes to equilibrium at a certain temperature, the mixture is found to contain 1.00 mol of SO_3. What is the amount of O_2 at equilibrium? a. 0.00 mol b. 1.00 mol c. 1.50 mol d. 2.00 mol e. 2.50 mol The value of K_eq for the equilibrium: H_2 + I_2 rightarrow 2HI = 794 is 794 at 25 degree C. At this temperature, what is the value of K_eq for the equilibrium: HI (g) doubleheadarrow 1/2 H_2 (g) + 1/2 I_2 (g) a. 00035 b. 0.0013 c. 28 d. 397 e. 1588

Explanation / Answer

(3)

Initial moles of SO2 = 2 mol

Moles of O2 = 2 mol

Moles of SO2 FINAL = 1 mol

From stochiometry ,

Initial - 2*x = final

2-2*x= 1

X = 0.5 moles

For oxygen

Final moles = initial -1*x = 2-1*0.5 = 1.5 moles

(4)

Here the reaction is reversed and halved.

If reaction is reversed kp new = 1/kp old

If the reaction is halved kp new = sqrt(kp old)

So in this condition,

Kp new = 1/sqrt(Kp old) = 1/794 = 0.035

Option (a)

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