Ch 5I question #19

Moderators: Chem_Mod, Chem_Admin

205192823
Posts: 100
Joined: Wed Sep 18, 2019 12:19 am

Ch 5I question #19

Postby 205192823 » Mon Feb 03, 2020 10:11 pm

A reaction mixture that consisted of 0.400 mol H2 and 1.60 mol I2 was introduced into a flask of volume 3.00 L and heated. At equilibrium, 60.0% of the hydrogen gas had reacted. What is the equilibrium constant K for the reaction H2(g)1I2(g) ∆ 2 HI(g) at this temperature?
When solving this do you have to assume that since only 60% of hydrogen gas has reacted (0.400 mol H2), would HI be equal to the 2(60% of H2 + 1.60 mol of I2)?

Benjamin Feng 1B
Posts: 102
Joined: Sat Sep 07, 2019 12:19 am

Re: Ch 5I question #19

Postby Benjamin Feng 1B » Mon Feb 03, 2020 11:35 pm

Yes, since for H2 to be used, HI must form (twice as much). No matter is created or destroyed, so it must have come from the reactants given.

Jessica Chen 2C
Posts: 103
Joined: Thu Jul 11, 2019 12:17 am

Re: Ch 5I question #19

Postby Jessica Chen 2C » Wed Feb 05, 2020 1:15 am

If you're asking if the final equilibrium value of HI is 2(60% of H2 + 1.60 mol of I2), then no, that's not entirely correct. When you set up an ICE table, you can see that for the "C(hange)" row, the change in the value of HI is +2*x because there are two moles. To find x, you use the fact that 60% of H2 has reacted to find the final equilibrium value of H2, and you subtract that from the initial concentration of H2 to find x.


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 1 guest