What is the value of kc for the decomposition of hi at 623 k?

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