Equilibrium Constant Calculation (1 Viewer)

[Menomaths]

At 900K 5M of H₂ gas, 4M of I₂ gas and 1.6M of hydrogen iodide were mixed together in a 1L container.

After equilibrium has been achieved, the vessel was found to contain 3.6 moles of hydrogen gas.

Calculate the value of equilibrium constant for this reaction at 900K.

Don't need working out at the moment, I just want to compare my answer.

 

[Menomaths]

One more. Can't understand the highlighted step

 

[RealiseNothing]

One more. Can't understand the highlighted step

There are 2 moles of iodine trichloride for every mole of iodine gas. Since iodine gas increased by +15, then iodine trichloride must decrease by twice that, i.e. -30.

 

[RealiseNothing]

At 900K 5M of H₂ gas, 4M of I₂ gas and 1.6M of hydrogen iodide were mixed together in a 1L container.

After equilibrium has been achieved, the vessel was found to contain 3.6 moles of hydrogen gas.

Calculate the value of equilibrium constant for this reaction at 900K.

Don't need working out at the moment, I just want to compare my answer.

Doing it quickly I got 2

 

[Menomaths]

There are 2 moles of iodine trichloride for every mole of iodine gas. Since iodine gas increased by +15, then iodine trichloride must decrease by twice that, i.e. -30.

Thanks

 

[Menomaths]

Doing it quickly I got 2

Can you post your working

Was your answer 1.9?(if it was no need to post up working)

 

[RealiseNothing]

H₂ + I₂ 2HI

Initially 5M of H₂ and 3.6M at the end. So a decrease by 1.4M.

The mole ratio is 1:1 for H₂:I₂ so I₂ also decreases by 1.4M to go from 4M initially to 2.6M at the end.

For H₂:2HI the mole ratio is 1:2 and so HI increases by 2.8M to go from 1.6M to 4.4M. Hence using the equilibrium constant equation:

 

[Menomaths]

Thanks