Original post by Mr Tall
4 mol of methanoic acid are reacted with 6 mol of ethanol
HCOOH(l) + c2h50h (l) <=> HCOOC2H5(l) + H2O(l)
the equilibrium mixture contains 3 mol of HCOOC2H5.
What is the equilibrium constant for this reaction?
thanks
I have no idea how to do this...
HCOOH(l) + c2h50h (l) <=> HCOOC2H5(l) + H2O(l)
the equilibrium mixture contains 3 mol of HCOOC2H5.
What is the equilibrium constant for this reaction?
thanks
I have no idea how to do this...This is basic stoichiometry.
You start off with 0 moles of ester, you end up with 3 mol of ester then the reaction has produced 3 moles of ester.
How many moles of alcohol and acid must react to produce 3 moles of ester?
How many moles from the original mixture are left behind unreacted?
Original post by charco
This is basic stoichiometry.
You start off with 0 moles of ester, you end up with 3 mol of ester then the reaction has produced 3 moles of ester.
How many moles of alcohol and acid must react to produce 3 moles of ester?
How many moles from the original mixture are left behind unreacted?
You start off with 0 moles of ester, you end up with 3 mol of ester then the reaction has produced 3 moles of ester.
How many moles of alcohol and acid must react to produce 3 moles of ester?
How many moles from the original mixture are left behind unreacted?
3 mol must react to give 3 mol of ester surely?
Original post by Mr Tall
3 mol must react to give 3 mol of ester surely?
... and so how much of each is left over?
I find it helps If you write out the equation and then put three rows below it and place the values underneath the equation for initial moles, moles reacted and moles at equilibrium
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(edited 10 years ago)
Original post by Mr Tall
4 mol of methanoic acid are reacted with 6 mol of ethanol
HCOOH(l) + c2h50h (l) <=> HCOOC2H5(l) + H2O(l)
the equilibrium mixture contains 3 mol of HCOOC2H5.
What is the equilibrium constant for this reaction?
thanks I have no idea how to do this...
HCOOH(l) + c2h50h (l) <=> HCOOC2H5(l) + H2O(l)
the equilibrium mixture contains 3 mol of HCOOC2H5.
What is the equilibrium constant for this reaction?
thanks
I have no idea how to do this...The stoichiometry is 1:1:1:1
So if 3 moles of ester are produced it would also produce 3 moles of water
3 moles of acid and 3 moles of alcohol will need to have reacted, so you can use this to work out the equilibrium moles of acid and alcohol, then put them into the kc equation thing
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