share
ChemistryHow many grams of each substance will be present at equilibrium in the reaction of ethanol and ethanoic acid?
[+6] [1] Jesse
[2014-03-07 08:17:46]
[ equilibrium esters ]
[ https://chemistry.stackexchange.com/questions/8978/how-many-grams-of-each-substance-will-be-present-at-equilibrium-in-the-reaction ]

In the following reaction, $K_c=4.0$: $$\ce{C2H5OH + CH3COOH <=> CH3COOC2H5 + H2O}$$

A reaction is allowed to occur in a mixture of $\pu{17.2 g}$ $\ce{C2H5OH}$, $\pu{23.8 g}$ $\ce{CH3COOH}$, $\pu{48.6 g}$ $\ce{CH3COOC2H5}$, and $\pu{71.2 g}$ $\ce{H2O}$

(a) In what direction will a net change occur?

(b) How many grams of each substance will be present at equilibrium?

This is what I did for part (a): $$Q_c = \frac{[\ce{CH3COOC2H5}][\ce{H2O}]}{[\ce{C2H5OH}][\ce{CH3COOH}]}$$

First, get the concentration of $\ce{C2H5OH}$, $\ce{CH3COOH}$, $\ce{CH3COOC2H5}$, and $\ce{H2O}$ using the given amount of each substance.

$$\begin{align} [\ce{C2H5OH}] &= \pu{17.2 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} & &= \pu{0.373 mol} \\ [\ce{CH3COOH}] &= \pu{23.8 g} \times \frac{\pu{1mol}}{\pu{60.1 g}} & &= \pu{0.396mol} \\ [\ce{CH3COOC2H5}] &= \pu{48.6 g} \times \frac{\pu{1 mol}}{\pu{88.1 g}} & &= \pu{0.552mol} \\ [\ce{H2O}] &= \pu{71.2 g} \times \frac{\pu{1 mol}}{\pu{18.0 g}} & &= \pu{3.96 mol} \\ \end{align}$$

Assuming the overall volume is $\pu{1 L}$, then $$Q_c = \frac{(\pu{0.552 M})(\pu{3.96 M})}{(\pu{0.373 M})(\pu{0.396 M})} = 14.8 $$

Since $K_c$ is $4.0$, and $Q_c$ is $14.84$, $Q_c$ is larger than $K_c$ and the direction of the reaction is to the left, towards the reactants.

I know how to solve for (a) but I have no idea how to get the amount of each substance at equilibrium for (b). The answer for (b) should be $\pu{68 g}$. I tried doing the ICE chart, but I did not get $\ce{68 g}$. Do I use the ICE chart for this part of the question or which method should I be using?

You say raction evolves to the left. Calculate by which amount concentrations vary until equilibrium is established. - mannaia
Do not understand why for each substance the mass should have the same value (68 g)... - mannaia
@mannaia Yep, it's only for water. The question should have asked specifically how many grams are produced when the ester is produced from the dehydration reaction. - user3735
I'll not give the correct answer yet, but the sum of grams of each substance in the final equations is different from the initial, attention to the mass conservation!!! At 25C and 1 ATM, the correct Kc for this reaction is 4.634, so what is the correct answer? - DanielCorrêa
First of all welcome to Chemistry SE. You're right there are mistakes in above answer, another member has further pointed it out. Second, no mention is given in the original question that working conditions are standard. So $K_{c}$ can have other values than the standard one. - mannaia
@mannaia Yep, I assumed this was occurring at $25^\circ\text{C}$ and 1 ATM. I resolved the calculation errors. Alternatively, we can use the van't Hoff isochore equation if the temperature is changed. - user3735
[+3] [2014-03-07 14:22:22] user3735 [ACCEPTED]

Knowing that $K_c=4.0$, equilibrium is achieved when $Q_c = K_c$.

Let's set up an ICE table and write down our equilibrium expression like you did:

Reactants:

$$\ce{C2H5OH + CH3COOH <=> CH3COOC2H5 + H2O}$$

\begin{array} {|c|c|c|c|c|} \hline \text{Initial conc.} & \pu{0.373 mol} & \pu{0.396 mol} & \pu{0.552 mol} & \pu{3.96 mol}\\ \hline \text{Change conc.} & -x & -x & +x & +x\\ \hline \text{End conc.} & \pu{0.373 mol} - x & \pu{0.396 mol} - x & \pu{0.552 mol} + x & \pu{3.96 mol} + x \\ \hline \end{array}

Assuming that our reaction is at equilibrium:

$$ K_c = \ce{\frac{[CH3COOC2H5][H2O]}{[C2H5OH][CH3COOH]}}$$

We plug in our values for our change in equilibrium:

$$ K_c = 4.0 = {\frac{[0.552 + x][3.96 + x]}{[0.373 - x][0.396 - x]}}$$

Solving for $x$, which is the change in concentration:

\begin{align} 4 &= \frac{x^2+4.512 x+2.18592}{x^2-0.769 x+0.147708}\\ 0 &= 3x^2-7.588 x-1.59509 \end{align}

Using the quadratic equation, or thankfully polysolv on modern calculators, we get two values for $x$:

\begin{align} x_1 &= 2.72449 \\ x_2 &= -0.195154 \end{align}

The negative zero is chosen, since the reaction will evolve to the left (being $Q_c>K_c$), thus increasing the concentration of the reactants. This will only be accounted for, if one takes the negative zero. If one chooses the other zero, one will go to the wrong opposite direction (towards a diminution of reactants concentration). In the particular case of the present exercise, one would even result in a negative concentration. Therefore:

\begin{align} \ce{[CH3COOH]} &= \pu{0.373 mol} - (\pu{-0.195154 mol}) & &= 0.568154\\ \ce{[C2H5OH]} &= \pu{0.396 mol} - (\pu{-0.195154 mol}) & &= 0.591154\\ \ce{[CH3COOC2H5]} &= \pu{0.552 mol} + (\pu{-0.195154 mol}) & &= 0.356846\\ \ce{[H2O]} &= \pu{3.96 mol} + (\pu{-0.195154 mol}) & &= 3.764846 \end{align}

Therefore:

\begin{align} \pu{0.568154 mol} \ \ce{CH3COOH} \times \frac{\pu{60.1 g}}{\pu{1 mol} \ \ce{CH3OOH}} &= \pu{34.1 g}\\ \pu{0.591154 mol} \ \ce{C2H5OH} \times \frac{\pu{46.1 g}}{\pu{1 mol} \ \ce{C2H5OH}} &= \pu{27.3 g}\\ \pu{0.356846 mol} \ \ce{CH3COOC2H5} \times \frac{\pu{88.1 g}}{\pu{1 mol} \ \ce{CH3COOC2H5}} &= \pu{31.4 g}\\ \pu{3.764846 mol} \ \ce{H2O} \times \frac{\pu{18.0 g}}{\pu{1 mol} \ \ce{H2O}} &= \pu{67.8 g} \end{align}

Initial product sum is $\pu{160.8 g}$.

Equilibrium product sum is $\pu{160.6 g}$. $\pu{0.2 g}$ is likely due to truncation error in calculations.

This problem is really cruel... whoever assigned it... :/


1