A2 Equilibrium Help!

Hi there i need help on how to do the following question:

At 723K, hydrogen and iodine react together and the following equilibrium is established: H₂(g) + I₂(g) == 2HI(g)
The value of K_c for this equilibrium is 64. In an experiment, equal amounts of hydrogen and iodine were mixed together, and the equilibrium mixture of the three gases in a container of volume 1 dm³ at 723K was found to contain 1.5 moles of iodine. Calculate the concentration of hydrogen iodide in the mixture at 723K.

Help would be really appreciated as i just don't know how to get to the answer, explanation as to how you did so will also be useful. Chemistry calculations has always been my downfall so help on this would be appreciated! :smile:

Reply 1

QuantumSuicide

If there's 1.5 mol iodine at equlibrium then there is 3 mol HI (from stoichiometry)

Then just apply n=cv
C=n/v
3/1
= 3 mol dm^-3

I is right?

Posted from TSR Mobile

(edited 9 years ago)

Original post by QuantumSuicide

If there's 1.5 mol iodine at equlibrium then there is 3 mol HI (from stoichiometry)

Then just apply n=cv
C=n/v
3/1
= 3 mol dm^-3

I is right?

Posted from TSR Mobile

no

Reply 3

TSR George

OP

17

Original post by QuantumSuicide

If there's 1.5 mol iodine at equlibrium then there is 3 mol HI (from stoichiometry)

Then just apply n=cv
C=n/v
3/1
= 3 mol dm^-3

I is right?

Posted from TSR Mobile

The answer is actually: 12moldm^3

I don't know how they actually got this. :frown:

Reply 4

QuantumSuicide

Original post by charco

no

Ooooh, i know how to do this.

Do kc = products over reactants, then solve it

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Reply 5

TSR George

OP

17

Original post by QuantumSuicide

Ooooh, i know how to do this.

Do kc = products over reactants, then solve it

Posted from TSR Mobile

Original post by charco

no

Can you show me please?

Original post by Jatyization

Hi there i need help on how to do the following question:

At 723K, hydrogen and iodine react together and the following equilibrium is established: H₂(g) + I₂(g) == 2HI(g)
The value of K_c for this equilibrium is 64. In an experiment, equal amounts of hydrogen and iodine were mixed together, and the equilibrium mixture of the three gases in a container of volume 1 dm³ at 723K was found to contain 1.5 moles of iodine. Calculate the concentration of hydrogen iodide in the mixture at 723K.

Help would be really appreciated as i just don't know how to get to the answer, explanation as to how you did so will also be useful. Chemistry calculations has always been my downfall so help on this would be appreciated! :smile:

There must also be the same number of moles of hydrogen (the iodine started off the same as hydrogen and equimolar amounts must have reacted from the equation stoichiometry)

Now you apply the equilibrium law:

k = [HI]²/

[I2]

64 = [HI]²/1.5*1.5

[HI]² = 64 * 2.25 = 144

[HI] = 12

Reply 7

Gibber96

9

Original post by Jatyization

Hi there i need help on how to do the following question:

At 723K, hydrogen and iodine react together and the following equilibrium is established: H₂(g) + I₂(g) == 2HI(g)
The value of K_c for this equilibrium is 64. In an experiment, equal amounts of hydrogen and iodine were mixed together, and the equilibrium mixture of the three gases in a container of volume 1 dm³ at 723K was found to contain 1.5 moles of iodine. Calculate the concentration of hydrogen iodide in the mixture at 723K.

Help would be really appreciated as i just don't know how to get to the answer, explanation as to how you did so will also be useful. Chemistry calculations has always been my downfall so help on this would be appreciated! :smile:

Right dude. It helps to do a little table with your three species and then their concentrations at equilibrium and at the start. So initially you have equal concs of H and I and no HI at all. At eq you have 1.5 moles I, which means you must have 1.5 moles of H. Moles HI is unknown at this point.

Using Kc = 64, you know that 64=x^2/1.5^2. I think you can leave the concentrations as moles as the volume is 1dm so the ratio will stay the same. If you then multiply 64 by 1.5^2 and square root the answer you get 12mol dm^-3 for the concentration of HI. I think that's right.

Reply 8

TSR George

OP

17

Original post by charco

There must also be the same number of moles of hydrogen (the iodine started off the same as hydrogen and equimolar amounts must have reacted from the equation stoichiometry)

Now you apply the equilibrium law:

k = [HI]²/

[I2]

64 = [HI]²/1.5*1.5

[HI]² = 64 * 2.25 = 144

[HI] = 12

Thank you so much for your help! I really do appreciate it and i understand this a lot more now. It was actually really easy, i got confused because i thought we must use algebra to work this out :redface: