HW 5I.3

[Amy Xiao 1I]

"In a gas-phase equilibrium mixture of H2, I2, and HI at 500. K, [HI] is 2.21 x 10^-3 mol/L and [I2] is 1.46 x 10^-3 mol/L. Given the value of the equilibrium constant in Table 5G.2, calculate the equilibrium molar concentration of H2."

How do you solve this? I looked at the table and I think I use the reaction H2(g) + I2(g) <-> 2HI(g) and for that reaction, K=160 at 500K. Should I be converting the concentrations to partial pressures because they are gases? If so, what R do I use? If not, what should the expression be to find the answer (2.1 x 10^-5 mol/L)?

[DesireBrown1J]

You do not have to convert them to partial pressure. You can use Kc=[R]/[P] to solve for this problem by plugging in Kc from the Table 5G.2 and [HI] is 2.21 x 10^-3 mol/L and [I2] is 1.46 x 10^-3 mol/L to solve for the equilibrium molar concentration of H2. If you completed 5I.1, it follows the same steps.

[Osvaldo SanchezF -1H]

So you do not need to convert to partial pressure though you can if you want but its a waste of time since they gave you the concentrations already. And knowing the equilibrium values of HI and I2 and the K value you don't have to do the ICE table and can just plug in the values to the equilibrium constant equation to find the [H2]. So it would be K=[HI]/[H2][I2] and with the values plunged in it would be 160=(2.21*10^-3)^2/(1.46*10^-3)*[H2]. So now just solve for [H2] and you should get the answer. you can use this method when all other values are given except one at equilibrium. You would have to use the ICE table however if the values given were not those found at equilibrium.