#5-8 If the equilibrium concentration in a 2.00L flask of N_2O_4 is 0.20 mol and

Question

#5-8

If the equilibrium concentration in a 2.00L flask of N_2O_4 is 0.20 mol and the equilibrium concentration of NO_2 is 0.0020 mol, find K. Equation given in Problem 1. For the following aqueous rxn, 0.15M C and 0.087M D are allowed to react. At equilibrium, only 0.009M D remains. Calculate K and of the equilibrium concentrations. 2A + B C + 3D The following gas phase reaction has K_e = 82.9. If we start with 0.062 mol of A in a 1.00L flask, what are the equilibrium concentrations? 2A B + C PC1_5 PCl_3 + Cl_2 At 250. degree C, 0.018 mol PCl_5, 0.030 mol PCl_3, and 0.040 mol Cl_2 are added to a 1.00 L flask. Which direction will the rxn proceed, and what are the equilibrium concentrations? K = 0.017

Explanation / Answer

1) number of moles = molarity (mol/L) * volume (L)

molarity of N2O4 = 0.2 mol / 2L = 0.1 mol/L

  molarity of NO2 = 0.002 mol / 2L = 0.001 mol/L

The equilibrium reaction is :

  N2O4 <----> 2NO2

Thus equilibrium constant = [NO2]2 /[N2O4]

   Kc = (0.001)2 / 0.1

Kc =0.00001

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