Question

1. Remember for calorimetry that sys - AH' and that you measured surr 2. Calculate AH', using the following equation: 4surr = -Asys 3. The equation for the dissolution of urea is: urea(s) + H2O(l) = urea(aq) 4. Because pure solids and pure liquids do not appear in the equilibrium expression, we get the following: Keg = [urea] 5. Calculate AG and AS 6. Is this an endothermic or exothermic process? 7. Is this a spontaneous process? 8. Explain, from a molecular level, your answers to the two previous questions.

This is a thermodynamics of urea dissolution lab. Temperature of water was 25.6 C Temperature of water mixed with urea was 21.7 C Mass of 25ml of water - 24.5 g Heat capacity of water given: c: 2.184 J/g C Mass of urea is 1.5 g Calculate q and please answer the questions above. Thanks.

Answer 1

2. We can calculate the heat of the surroundings taking into account the following equation:

$q=m\cdot c\cdot \Delta T$

Where q is the heat, m is the mass of the system, c is the heat capacity and delta T is the change in temperature (the difference between the final and initial temperatures). I'm going to correct your value for the heat capacity of water, which is 4.184 J/(g°C) and use it:

$q=24.5g\cdot 4.184\frac{J}{g\cdot ^{o}C}\cdot (21.7-25.6)^{o}C=-400J$

This means that the heat of the system was 400 J, which is equal to ∆H°.

4. The value for K_eq = [urea] is given by (this assumption is pretty borderline, since we are assuming all urea has dissolved to calculate the concentration, which is incoherent if we consider that there is an equilibrium between dissolved and non-dissolved urea, we'll go on in order to show the required calculations):

$[urea]=\frac{\frac{m}{m_{molar}}}{V(L)}=\frac{\frac{1.5g}{60.06g/mol}}{0.025L}=0.999M$

5. The value of ∆G° can be calculated using (considering the initial temperature of 25.6 °C):

$\Delta G^{o}=-RTln(K)=-8.314\frac{J}{mol\cdot K}\cdot 298.6K\cdot ln(0.999)=2.48\frac{J}{mol}$

And now ∆S° can be calculated using:

$\Delta G^{o}=\Delta H^{o}-T\Delta S^{o}$

Which can be rearranged to:

$\Delta S^{o}=-\frac{\Delta G^{o}-\Delta H^{o}}{T}=-\frac{2.48J/mol-400J/mol}{298.6K}=1.33\frac{J}{mol\cdot K}$

6. This is an endothermic process, since ∆H° has a positive value, which means that the system "absorbs" heat.

7. The positive ∆G° value tells us that this is not a spontaneous process.