Question


Consider the reaction of 48.3 mL of 1.0 M HCI and 49.2 mL of 1.0 M NaOH. Reaction caused the initial temperature to increase from 22.4 °C to 28.7 °C. 2. a. Determine the heat change for this reaction.

B. Determine \Delta H for the reaction in KJ/mol of water formed.

C. Assuming the heat absorbed or released by the calorimeter is nonegligible, how does ignoring the heat absorbed or released by the calorimeter affect the magnitude of \Delta H? Explain your answer.

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Answer #1

2)

total volume = 48.3 + 49.2 = 97.5 mL

mass of solution = 97.5 g

temperature rise = 28.7 - 22.4 = 6.3 oC

specific heat = 4.184 J / g oC

Q = m Cp dT

   = 97.5 x 4.184 x 6.3

Q = 2570 J

heat change for this reaction = 2570 J

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