Question

Propy alchol 6. Give the following values in kJ/mol for CH,(CH,);CH,OH at 1 atm and 25°C: Show your work below for full credit. ΔΕ- ΔΗ- W =

What i have so far .

Got stuck on #6

For CH3(CH2)2CH2OH (propyl alchol)

Answer 1

volume of sample 1.00mL Density of Sample = Blokg/m3 = 810x10°9 og = 0.810 g/mL Initial temperature → 21.00% 10mL Final temperature * 23.81°C to CH3 CH2 CH2 CH2OH (1) + 6 02 (9) 4CO₂(g) + 5H20 (1) 2. Mass of the 1-Butanol = Volume x densily B = 1 ml x 0.810 g/mL = 0+810 9 3. Heat released by the combustion of 0.8109 of 1-butanol = heat absorbed by the calorimeter = Heat capacity of Calorimeterx temperature change = 10:3 (KJ/°c) (23.81–21.00) °c = 10.3 x 2.81 KJ = 28.943 KJ I mot I-butanol releases heat during combustion = 281943 x 74.1216 KJ/mol 0810 = 2648.52KJhent Molar mass of l-butanol 1 = 74.1216 g/mol - Heat of combustion (4He) for 1-butanol =-2648.52KJ/mol 4. As the heat of combustion is measured using a bomb- calorimeter (constant volume) is it equal to both 4H and AE because w = 4E 50, 4E = AH =-2648.52 KJ/mol. 5. As Volume is constant work done (W --PAV) = 0 6. CH3(CH2)2CH2OH → 1-Butanol (CH3CH2CH2CH3OH) AE =-2648.52 KJ/mol. AH = - 2648:52 KJ/mol N