Equilibrium in reaction of C2H5CO2H(l) + C3H7OH(l)

In summary, the given reaction between propanoic acid and n-propanol produces n-propyl propanoate and water when heated and distilled. At room temperature, the equilibrium constant is approximately 3. To calculate the number of moles of reactants and products in the equilibrium mixture, you can use the equilibrium equation and substitute in the equilibrium concentrations of each substance, divided by the total reaction volume. Using the data given, including the boiling point of n-propyl propanoate, volume of water collected, and weight of the product, you can estimate the concentrations of all substances at a particular temperature and solve for the number of moles of water at equilibrium. However, there may be some ambiguity in the question regarding
  • #1
Mitchtwitchita
190
0
C2H5CO2H(l) + C3H7OH(l) ---> C2H5COOC3H7(l) + H2O(l)

18.5 mL propanoic acid are mixed with 19.0 mL n-propanol to produce n-propyl propanoate when heated and distilled.

At room temperature, the equilibrium constant, Keq is approximately 3. Calculate the number of moles of reactants and products expected in the equilibrium mixture of this reaction. (Hint: Let X be the number of moles of each product formed at equilibrium. Substitute into the equilibrium equation using the moles of each reactant and product at equilibrium each divided by the total reaction volume. Leave the equilibrium concentrations as fractions. You can then take the square root of both sides of the equation after canceling like terms, i.e. the volumes.)

Other Data:

Boiling Point of n-propyl propanoate = 122 degrees C
Volume of water collected = 3.31 mL
Weight of product (C2H5COOC3H7) = 7.4846 g

I have no idea of how to start this problem, can anybody give me a push in the right direction please?
 
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  • #2
I can give you a push in what I think is the right direction but the question
as stated is not clear to me. From the information given, you can easily
estimate the concentrations of all substances in the mixture at a particular
temperature. i.e. if the mixture were refluxed at some temperature T:
Let n(p) be the number of moles of water at equilibrium at temperature T.
Then the number of moles of ester is also n(p), the number of moles of
acid and alcohol is x(ac) = n(ac) - n(p) and x(al) = n(al) - n(p), where
n(ac) and n(al) are the number of moles of acid and alcohol, repsectively,
that are available (given as 18.5 mL and 19 mL respectively---to compute
the number of moles n(ac) and n(al) you need the density
at room temperature and the molecular weights). Given any temperature T,
you first need to estimate the equilibrium constant at that T, K(T).

d(log K)/d(1/T) = -Delta HoR ,

where "log" is the natural logarithm, Usually, one can assume that the
enthalpy of reaction is constant over the temperature range; if this is
not true, then you must express it as a function of T as well. Then,
obtain n(p) by equating the equilibrium-constant-expression to K(T) and
solving for the one unknown, n(p). That's it.

But maybe I have not interpreted the question properly. First of all,
I don't understand the reference to "volumes" in the hint. There is
no big volume change on reaction, even in the gas phase, so volume should
play a negligible role. Secondly, the question mentions distillation.
Water and n-propyl alcohol both have boiling points near 100 C,
whereas propanoic acid boils at about 141 C. Removal of water would
drive the equilibrium in the direction of products but removal of
alcohol would have the opposite effect.

Finally, the number of moles of water collected is about 2.85 times the
number of moles of product (propyl ester), but these should be equal.
Was the entire mixture distilled? If not, there must be a considerable
amount of product that was not distilled, or there are other reactions
going on that were not mentioned.

I conclude that I have misunderstood the question or have made some
mistake in my rough calculations. Nevertheless, maybe you will get
some idea of how to approach this better, and that would also
constitute a "push".
 
  • #3


I would approach this problem by first understanding the chemical reaction that is occurring. The reaction is an esterification reaction, where propanoic acid (C2H5CO2H) reacts with n-propanol (C3H7OH) to form n-propyl propanoate (C2H5COOC3H7) and water (H2O). This reaction is reversible, meaning that the products can also react to form the reactants.

The equilibrium constant (Keq) is a measure of the ratio of products to reactants at equilibrium. In this case, a Keq of 3 means that there are 3 moles of products for every 1 mole of reactants at equilibrium. Using this information, we can set up an equilibrium expression:

Keq = [C2H5COOC3H7] / [C2H5CO2H] [C3H7OH]

Where [C2H5COOC3H7], [C2H5CO2H], and [C3H7OH] represent the equilibrium concentrations of each compound. Since we are given the volumes of each reactant and product, we can calculate their molarities using the formula Molarity = moles / volume. We can also assume that at equilibrium, the number of moles of each product formed (X) is equal to the number of moles of each reactant consumed (X). This is because the reaction is in a closed system and the total number of moles of each compound remains constant.

Substituting these values into the equilibrium expression, we get:

3 = X / (18.5 mL + 7.4846 g / 74.08 g/mol) X / (19.0 mL + 3.31 mL)

Solving for X, we get X = 0.0666 moles.

This means that at equilibrium, there will be 0.0666 moles of each product formed and 0.0666 moles of each reactant consumed. To calculate the mass of each compound at equilibrium, we can use the molar mass of each compound. For example, the mass of n-propyl propanoate (C2H5COOC3H7) at equilibrium would be:

Mass = 0.0666 moles x 74.08 g/mol = 4.93 g

Similarly
 

Related to Equilibrium in reaction of C2H5CO2H(l) + C3H7OH(l)

1. What is the equilibrium constant for the reaction between C2H5CO2H(l) and C3H7OH(l)?

The equilibrium constant for this reaction is equal to the concentration of the products (C2H5CO2H and C3H7OH) divided by the concentration of the reactants (C2H5CO2H and C3H7OH). This value is dependent on temperature and can be calculated using the equilibrium expression: Kc = [C2H5CO2H][C3H7OH] / [C2H5CO2H][C3H7OH].

2. How does temperature affect the equilibrium of this reaction?

According to Le Chatelier's principle, an increase in temperature will favor the endothermic reaction in this exothermic reaction. This means that more products will be formed at higher temperatures, resulting in a higher equilibrium constant. Conversely, a decrease in temperature will favor the exothermic reaction, resulting in a lower equilibrium constant.

3. Is the reaction between C2H5CO2H(l) and C3H7OH(l) exothermic or endothermic?

This reaction is exothermic, meaning that it releases heat during the reaction. This is due to the formation of new bonds between the products, which releases energy.

4. How can the reaction be shifted to favor the formation of more products?

According to Le Chatelier's principle, increasing the concentration of the reactants (C2H5CO2H and C3H7OH) or decreasing the concentration of the products (C2H5CO2H and C3H7OH) will shift the equilibrium to the right, favoring the formation of more products. Additionally, lowering the temperature or increasing the pressure can also shift the equilibrium to the right.

5. Can the equilibrium constant for this reaction be changed?

The equilibrium constant for this reaction is dependent on temperature and cannot be changed unless the temperature is changed. However, the position of the equilibrium can be shifted by changing the concentration of the reactants or products, as well as the temperature or pressure. This will result in a new equilibrium constant for the reaction.

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