1.) Consider the reaction Br 2 (g) + Cl 2 (g) <===> 2BrCl(g) Calculate the value

Question

1.) Consider the reaction

Br2(g) + Cl2(g) <===> 2BrCl(g)

Calculate the value of Kp at 400?C if the partial pressures of BrCl, Br2, and Cl2 are 3.74 atm, 2.00 atm, and 1.00 atm, respectively when the system is at equilibrium.

2.) Consider the reactions

2NO + O2 <===> 2 NO2      K = a

2 NO2 <===> N2O4             K = b

The value of the equilibrium constant for the reaction 4NO + 2O2 <===> 2N2O4 is

3.) A student made a reaction mixture as directed in the lab manual, except he forgot to add the catalyst. If everyone's experiment reached equilibrium, how was that student's experiment different?

Explanation / Answer

Q.1: Br2(g) + Cl2(g) <===> 2BrCl(g)

For the above re]action

Kp = [PBrCl,g]2 / [PBr2,g]x[PCl2,g]

=> Kp = (3.74 atm)2 / (2.00 atm)x(1.00 atm)

=> Kp = 7.0 (answer)

Hence option A is correct

Q.2: The given equilibrium reactions are

2NO + O2 <===> 2 NO2      K = a ----- (1)

2 NO2 <===> N2O4             K = b -----(2)

Adding the above two reaction we get

2NO + O2 + 2NO2 <=== > 2NO2 + N2O4 or

2NO + O2 <=====> N2O4 : K = a x b ------ (3)

Now multiplying the above reaction (3) by 2 we get

4NO + 2O2 <=====> 2N2O4 : K = (a x b)x(axb) = (ab)2  (answer)

Hence option D is correct.

Q.3: (D) The one without a catalyst took longer to reach equilibrium, but the amount of product was the same as in the other experiments.

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