Design and carry out an experiment to determine the equilibrium constant, Keq, for this reaction at — Chemical Equilibrium Chemistry Question
Problem Context
Design and carry out an experiment to determine the equilibrium constant, Keq, for this reaction at room temperature.
urea(s) + H2O(l) æ urea(aq)
You will be asked to describe the method you developed to solve this problem.
Given: molar mass of urea, CO(NH2)2 = 60.0 g·mol–1
molarity of pure H2O = 55.5 mol·L–1
Give a brief description of your experimental plan.
Model Answer
A good plan recognized that it was necessary to determine how much water was required to completely dissolve 4.0 g of urea. For example, a good plan might consist of these steps.
1) Add water in small increments to 4.0 g of urea in the graduated centrifuge tube.
2) Cap the tube and shake after each addition.
3) If any solid remains, add another small portion of water. Cap and shake the tube.
4) Continue to add water until all the urea is dissolved.
5) Record the total volume of solution and/or the total volume of water added.*
6) Repeat with the second 4.0 g sample of urea.
An average plan was either missing one of these components or had less detail in two or more of these components.
A weak plan had minimal detail about how the experiment would be conducted.
Record your data and other observations.
Model Answer
Sample Data
Total Volume of Solution* Mass Urea
Trial 1 7.3 mL 4.0 g
Trial 2 7.5 mL 4.0 g
Many students also observed that as the urea dissolved, the tube felt cool to the touch. Some allowed time for the tube to return to room temperature before making final observations of volume.
What value did you calculate for the equilibrium constant? Show your methods clearly.
Model Answer
Sample calculations for Trial 1
1) Moles of urea = 4.0 g urea × 1 mol urea / 60.0 g urea = 0.067 mol urea
2) Molarity of urea solution = 0.067 mol urea / 0.0073 L solution = 9.2 M
3) Calculation of Keq if assume that the concentration of water is in standard state.
urea(s) + H2O(l) æ urea(aq)
Keq = Keq = 9.2
* A superior plan recognized that because there is a high concentration of urea in the saturated solution, the solution cannot be treated as a dilute solution in which the concentration of the solvent is the same as that of pure water. Points were awarded to students who realized this and adjusted their experimental approach. This requires knowing the volume of water added, not just the volume of the resulting solution. For example, if 4.3 mL of water was added, the final volume of the solution was reported as 7.3 mL. The molarity of water in the saturated solution can then be calculated, as shown in this example.
4.3 mL H2O / 7.3 mL solution × 55 M = 32 M and Keq = / [H2O(l)] and Keq = (9.2 M) / (1)(32 M / 55.5 M) = 16
Some students elected to obtain more samples by dividing the given mass of urea. The mass of smaller samples of urea was estimated by calculating the density of urea, given the known mass and the volume markings on the graduated centrifuge tube.