The following table contains data for the equilibrium reaction
CH3COOH(g)+ C2H5OH(g)↔ CH3COOC2H5(g)+ H2O(g)
T = 100oC.
Each row in the table represents a different experiment (diffferent intial concentrations).
Initial concentration Equilibrium concentration
--------------------------- -------------------------
[CH3COOH(g)] [C2H5OH(g)] [CH3COOC2H5(g)]
mol/L mol/L mol/L
---------------------------------------------------------------
1.00 (not given) 0.166
1.00 (not given) 0.616
1.00 (not given) 0.967
What is [H2O(g)] at equilibrium in the third experiment (third row of values) ?
HINT 1 : You solve this problem by inspection of the table values. It is helpful to write out the reaction and note the changes in reactants and products, in terms of x. Then fill in one or more values of x that are given above. But you do not need to do any calculations.
HINT 2 : The initial concentrations of the products are each zero mole/L .
Enter a numeric answer only (no units)
It's tough to answer these tabular questions on the board because of the spacing problem. Because of the spacing problem I'm not sure I understand the problem. You say you have initial concn and equilibrium concn which is two columns, yet you list three values in the first experiment. I assume those are 1.00 for CH3COOH, not given for C2H5OH and 0.166 for CH3COOC2H5. But are those equilibrium or concns or initial concns?
You might do this.
Initial concns:
CH3COOH = ??
C2H5OH = ??
CH3COOC2H5 = ??
equilibrium concns:
CH3COOH = ??
C2H4OH = ??
CH3COOC2H5 = ??
Perhaps I can follow it then.
The first two columns are initial concentrations of [CH3COOH(g)]and [C2H5OH(g)]and the third is equilibrium concentration of [CH3COOC2H5(g)].
Each row is a different experiment so that's why there are 3 sets of values.
To find the equilibrium concentration of H2O in the third experiment, we need to examine the changes in reactants and products and use the given values from the table.
From the balanced reaction equation:
CH3COOH(g) + C2H5OH(g) ↔ CH3COOC2H5(g) + H2O(g)
We can see that one molecule of H2O is produced for every molecule of CH3COOC2H5 formed. Therefore, the equilibrium concentration of H2O will be equal to the concentration of CH3COOC2H5.
Looking at the third row of the table, the equilibrium concentration of CH3COOC2H5 is given as 0.967 mol/L. Therefore, the equilibrium concentration of H2O in the third experiment is also 0.967 mol/L.
So, [H2O(g)] at equilibrium in the third experiment is 0.967.