Question #6 Trial #1 Trial #2 Mass of urea 3.004 g 3.010 g Tinitial 24.0C 24.0C
ID: 925899 • Letter: Q
Question
Question #6
Trial #1 Trial #2
Mass of urea 3.004 g 3.010 g
Tinitial 24.0C 24.0C
Tfinal 20.1C 19.7C
In this question you will use your data from Lab 12 Question 3 in the table above to determine H soln (kJ/mol) for urea (CH4N2O).
Part 1: Determine the total heat involved in dissolving your samples of urea in 50.0 mL of water. Pay attention to units and significant figures. ,
Total heat involved Trial 1 = 8.6e2J and Trial 2 = 9.5e2
Part 2: Determine H soln in terms of kJ/mol for urea for both trials.
Trial #1 H soln kJ/mol:
Trial #2 H soln kJ/mol:
Please show me the calculation for Part 2. I know that molar mass of urea is 60.06.
Explanation / Answer
A model to calculate:
Trial 1:
Heat absorbed during dissolution of 3.004 g urea was
Q = mcdT = 53 g solution x 4.18 J/g.ºC x (20.1-24.0) = 860 J (heat is absorbed)
If the reaction is endothermic dH is positive
dH = - Q
3.004 g urea/ 60.06 g/mol = 0.050 mol
dH = 860 J/0.05 mol = 17 kJ/mol
Trial 2
dH = 950 J/0.05 mol= 19 kJ/mol