During an experiment, 3.0g of sodium acetate, NaC2H3O2, were added to 20.0 ml water at 20.0ºC. The temperature dropped to 18.0º

Question

During an experiment, 3.0g of sodium acetate, NaC₂H₃O₂, were added to 20.0 ml water at 20.0ºC.  The temperature dropped to 18.0ºC.

a) Is the reaction endothermic or exothermic? Explain.

b) Find the heat of reaction for sodium acetate dissolving in water in kJ/mol. (assume the specific heat of the solution is the same as that of water)

c) Draw a sketch of an energy vs. time graph representing this reaction.  Label the heat of reaction you calculated in b) on your diagram.

Answer 1

a) Since the temperature dropped, this tells us that the reaction is endothermic

b) heat of reaction = ∆H (or q) = mC∆T

m = mass of solution = 20 g + 3 g = 23 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 20º - 18º = 2º

q = (23 g)(4.184 J/gº)(2º)

q = 192.5 J (this is the heat for 3.0 g of NaC2H3O2)

To find q (∆H) per mole, use the molar mass of sodium acetate of 82.0 g/mole

3.0 g NaC2H3O2 x 1 mol / 82.0 g = 0.0366 moles

∆H/mole = 192.5 J / 0.0366 moles = 5260 J/mole = 5.26 kJ/mol = 5.3 kJ/mol (2 sig. figs.)

c) Not able to draw on this platform but just do an internet search for "energy diagram for an endothermic reaction" and you'll see what it looks like. The beginning of the curve will be lower on the y axis than the end of the curve. The difference on the y axis between these two points will be the heat of reaction on the diagram.