Kc questions
Could someone help me with this question, I've had a go but im sure there is another step that I've missed out.
A mixture of 1.90 mol of hydrogen and 1.90 mol of iodine was allowed to reach equilibrium at 710k. the equilibrium mixture was found to contain 3.00 mol of hydrogen iodide. calculate the equilibrium constant at 710k for the reaction.
H2 (g) + I2 (g) <-> 2HI (g)
My working out:
Kc = [2HI]^2 / [I2]
kc = [ 3.00] ^2 / [1.90] [ 1.90]
[6.00] / [1.90] [ 1.90]
= 1.66
A mixture of 1.90 mol of hydrogen and 1.90 mol of iodine was allowed to reach equilibrium at 710k. the equilibrium mixture was found to contain 3.00 mol of hydrogen iodide. calculate the equilibrium constant at 710k for the reaction.
H2 (g) + I2 (g) <-> 2HI (g)
My working out:
Kc = [2HI]^2 / [I2]
kc = [ 3.00] ^2 / [1.90] [ 1.90]
[6.00] / [1.90] [ 1.90]
= 1.66
gerrard1892
Could someone help me with this question, I've had a go but im sure there is another step that I've missed out.
A mixture of 1.90 mol of hydrogen and 1.90 mol of iodine was allowed to reach equilibrium at 710k. the equilibrium mixture was found to contain 3.00 mol of hydrogen iodide. calculate the equilibrium constant at 710k for the reaction.
H2 (g) + I2 (g) <-> 2HI (g)
My working out:
Kc = [2HI]^2 / [I2]
kc = [ 3.00] ^2 / [1.90] [ 1.90]
[6.00] / [1.90] [ 1.90]
= 1.66
A mixture of 1.90 mol of hydrogen and 1.90 mol of iodine was allowed to reach equilibrium at 710k. the equilibrium mixture was found to contain 3.00 mol of hydrogen iodide. calculate the equilibrium constant at 710k for the reaction.
H2 (g) + I2 (g) <-> 2HI (g)
My working out:
Kc = [2HI]^2 / [I2]
kc = [ 3.00] ^2 / [1.90] [ 1.90]
[6.00] / [1.90] [ 1.90]
= 1.66
Initial moles:
H2: 1.9, I2: 1.9, HI:0
Equilibrium moles:
H2: 0.4, I2: 0.4, HI:3.0
Kc = [HI]2/[I2] = 9/0.16 = 56.25
charco
Initial moles:
H2: 1.9, I2: 1.9, HI:0
Equilibrium moles:
H2: 0.4, I2: 0.4, HI:3.0
Kc = [HI]2/[I2] = 9/0.16 = 56.25
H2: 1.9, I2: 1.9, HI:0
Equilibrium moles:
H2: 0.4, I2: 0.4, HI:3.0
Kc = [HI]2/[I2] = 9/0.16 = 56.25
How did you get 0.4 moles for both H2 and I2?
gerrard1892
How did you get 0.4 moles for both H2 and I2?
You are told that the equilibrium moles of HI = 3.0
For this to happen 1.5 moles of both H2 and I2 have to react together (look at the stoichiometry of the reaction)
If 1.5 moles react out of 1.9 then 0.4 are left behind...
I understand thought that the HI had to be doubled because it was only one mole.
Could you help me with this part of another question there's no (g) for H20 do I just leave H20 out ?
gerrard1892
Could you help me with this part of another question there's no (g) for H20 do I just leave H20 out ?
clear as mud...
would you like me to guess the question maybe?
Sorry ill write the question out
If a mixture of 6.0g ethanoic acid and 6.9 of ethanol is allowed to reach equilibrium, 7.0 g of ethyl ethanoate is formed. calculate Kc?
CH3CO2H + C2H5OH <-> CH3CO2C2H5 + H20
I can see that you have to turn the (g) in to moles by the multiplying by the Mr. What about the H20?
CH3CO2H + C2H5OH <-> CH3CO2C2H5 + H20
I can see that you have to turn the (g) in to moles by the multiplying by the Mr. What about the H20?
gerrard1892
If a mixture of 6.0g ethanoic acid and 6.9 of ethanol is allowed to reach equilibrium, 7.0 g of ethyl ethanoate is formed. calculate Kc?
CH3CO2H + C2H5OH <-> CH3CO2C2H5 + H20
I can see that you have to turn the (g) in to moles by the multiplying by the Mr. What about the H20?
CH3CO2H + C2H5OH <-> CH3CO2C2H5 + H20
I can see that you have to turn the (g) in to moles by the multiplying by the Mr. What about the H20?
In this case the water is not a solvent, but a reagent, so you must include it.
charco
In this case the water is not a solvent, but a reagent, so you must include it.
The question does not have any (g) for H20. What would I use ?
gerrard1892
The question does not have any (g) for H20. What would I use ?
There are two approaches, both of which lead to the same answer:
1. The law of conservation of matter
2. The moles equivalent from the other reactants
Go on, have a go yourself...
Original post by gerrard1892
If a mixture of 6.0g ethanoic acid and 6.9 of ethanol is allowed to reach equilibrium, 7.0 g of ethyl ethanoate is formed. calculate Kc?
CH3CO2H + C2H5OH <-> CH3CO2C2H5 + H20
I can see that you have to turn the (g) in to moles by the multiplying by the Mr. What about the H20?
CH3CO2H + C2H5OH <-> CH3CO2C2H5 + H20
I can see that you have to turn the (g) in to moles by the multiplying by the Mr. What about the H20?
Let me have it solved from you
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