Question

# For an equilibrium reaction 2 A + 3 B ↔ 2 C, the equilibrium constant Kc...

For an equilibrium reaction 2 A + 3 B ↔ 2 C, the equilibrium constant Kc = 1.6×103.

For the reaction C ↔ A + 3/2 B, the value of equilibrium constant, Kc', is [Y]. (Fill in the blank. Show the value only. Report with proper number of significant figures and do not use scientific notation.)

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4. A sealed 1.0 L flask is charged with 0.500 mol of I2 and 0.500 mol of Br2. An equilibrium reaction ensues:

I2(g) + Br2(g) ↔ 2 IBr(g).

When the reaction reaches equilibrium, the flask contains 0.840 mol of IBr. What is the value of the equilibrium constant Kc, for this reaction?

 1 6.1 2 4 3 11 4 2.8 5 110

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5. Given a gas-phase reaction H2(g) + I2(g) ↔ 2 HI(g). The equilibrium constant of this reaction Kp = 0.654. At a certain moment of the reaction run, the partial pressures of gases in the reaction mixture are 0.127 atm for H2, 0.134 atm for I2, and 1.055 atm for HI. Calculate the Qp and answer the following questions.

The value of Qp is . (Fill in the blank; show the proper number of significant figures.)

Is the reaction at equilibrium?  (Fill in the blank indicating Yes or No only.)

If your answer to the above question is Yes, skip this question. If your answer is No, then in which direction does the reaction proceed?   (Fill in the blank indicating Left to righ or Right to left only.)

for 2A + 3B = 2C

Kc = [C]2/ [A]2[B]3

for C = A + 3/2 B

Kc' = [A][B]3/2/ [C] = (1/Kc)1/2 = (1/1.6×103 )1/2 = 0.025

4. As per the reaction, one mol of I2 reacts with one mol of Br2 to give two mol of IBr

therefore 0.420 mol of I2 reacted with 0.420 mol of Br2 to give 0.840 mol of IBr

therefore at equilibrium moles of I2 = moles of Br2 = 0.500 - 0.420 = 0.080

therefore Kc = (0.840)2/ 0.080 x 0.080 = 110.25. So the answer is the 5th option i.e. 110.

5. Qp = [HI]2/ [H2][I2] = (1.055)2/ 0.127 x 0.134 = 65.40

No, the reaction is not in equilibrium as Qp is higher than Kp.

The reaction will proceed towards left