percentage reacted

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Savannah Mance 4G
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Joined: Fri Aug 30, 2019 12:17 am

percentage reacted

Postby Savannah Mance 4G » Thu Jan 16, 2020 11:59 am

In the homework problem 5I.19 it asks for the equilibrium constant for the reaction of H2+I2<--->2HI if initially there is 0.4 mol of H2 and 1.6 mol of I2 in a 3.0 L container. They say that 60% of the hydrogen has reacted. When you are doing the ice table for the problem I thought that since the hydrogen is 60% reacted, I should take 60% of the mols of H2 before calculating the concentration but instead you're supposed to take 60% of the initial concentration that you calculate. I was wondering why you would take 60% of the concentration rather than the mol.

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Re: percentage reacted

Postby VPatankar_2L » Thu Jan 16, 2020 1:13 pm

When you are writing out your ICE table, you want to find the concentration of reactants and products at equilibrium. For your initial concentration, you would calculate 0.4 mol/3 L H2 to find the initial concentration of H2 which is 0.133 mol/L. For the change in concentration, you would multiply 0.133 mol/L by 0.6 since only 60% of H2 reacts, to get 0.08 mol/L. You are going to use the concentration instead of the mols because the number of moles of solute stay the same in a reaction.

Nick Fiorentino 1E
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Joined: Wed Sep 18, 2019 12:16 am

Re: percentage reacted

Postby Nick Fiorentino 1E » Thu Jan 16, 2020 3:53 pm

To find x, you would set .133-x (the statement for the final H2 concentration) equal to .40 x .133mol/L because 60% of the H2 was used, so 40% remains

Michelle Song 1I
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Re: percentage reacted

Postby Michelle Song 1I » Thu Jan 16, 2020 4:51 pm

Equilibrium constants are calculated using concentrations (or partial pressures, depending), so you could find the number of moles of reactants and product at equilibrium for this reaction, but you would have to convert them to concentrations in order to find the equilibrium constant.

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