Plus minus Experimental Determination of an Equilibrium Constant The decompositi

Question

Plus minus Experimental Determination of an Equilibrium Constant The decomposition of HI(g) is represented by the equation 2HI(g) equivalent H_2(g) + I_2(g) The following experiment was devised to determine the equilibrium constant of the reaction. HI(g) is introduced into five identical 400-cm^3 glass bulbs, and the five bulbs are maintained at 623 K.The amount of I_2 produced over time is measured by opening each bulb and titrating the contents with 0.0150 M Na_2S_2O_3(aq). The reaction of I_2 with the lit rant is I_2 + 2Na_2S_2O_3 right arrow Na_2S_4 O_6 + 2NaI Experimental data In which bulb would you expect the composition of gases to be closest to equilibrium?

Explanation / Answer

Answer is bulb 5.

I2 + 2Na2(S2O3) Na2(S4O6) + 2NaI

(28.68ml)(0.015 mmol/ml) = 0.4303 mmol [S2O3]2– consumed

Therefore, there were 0.4303/2 mmol = 0.2151 mmol I2 present

initial 0.280g HI = 280mg/(127.91 mg/mmol) = 2.189 mmol

.....2HI(g) H2(g) + I2(g)

2.189-2x........x.........x

Therefore, final HI = 2.189 -2(0.2151) mmol

Kc = (0.2151)^2/(2.189 - 2(0.2151))^2 = 0.01496

Because the # of moles of reactants and products are equal, the volume of the bulb isn't needed (it cancels out of Kc). For the same reason Kc = Kp

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