Question

Measuring Enthalpy Results

NaHCO₃(s) + HC₂H₃O₂(aq) à NaC₂H₃O₂(aq) + H₂O(l) + CO₂(g)     

	Trial 1	Trial 2
mass of baking soda(NaHCO3) added	5.4g	5.4g
mass of vinegar added	20ml	20ml
Final temperature, tf	16.7C	16.7°C
Initial temperature, ti	21.1°C	21.1°C
Change in temperature, ∆t	4.4°C	4.4°C
q (water) = mwCwTw	467.6 J	467.6 J
q(reaction)	-467.6 J	-467.6 J
moles of NaHCO3	.0642	.0642
ΔH (kJ/mol)	-7.27448 kJ/mol	-7.27448 kJ/mol

Complete questions using data from table and the answer from question 1

Q.1 –
ANSWER = 31.4 kJ/mol

Q.4 What are some errors in your setup that could have caused this error? What pieces of equipment would result in a better final heat measurement? Be specific.

Answer 1

Solution. The given reaction is NaHCO3 is) + HC2 H3 Oz (aq) - Not cag) to ea (2 H302 ! + H2014) + CO2(g). for therotical enthalpy we have equation Horn = { Hf Product - {He reactor Hf values in KJ/mol. OHI of Ha C2H₂O2 (aq) = - 486 kJ/mol A Hf of NaHCO3 (8) a go.8 KJ / mal A Hof of Nat laa) = -240.1 kJ/mol- A Hf of C2 H202 car a + 4.80 kJ) mal AHof of CO2 lg) - 289. 8 kJ substituting relves in eah we get. Han = ED Hip product - & Hf reactant. = 14-240)) +(-486) + (-393 5) +(-289.0) - [(-486) + (- 9503)] kJ/mol = [-1405.4 - (- 14368) KJ mal Johann è 31.4 kJ/moy.

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