A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution in the calorimeter (q =3.50 kJ), resulting in a temperature rise of 7.32C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49C. What is the change in the internal energy of the neutralization reaction?

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thomasdecabooter thomasdecabooter · Answer 1 · 2019-11-05T23:09:24+01:00

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

Step 1: Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

Step 2: Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

Step 3: Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

Step 4: Calculate change in internal energy

ΔU = Q + W W = 0 (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

lhmarianateixeira lhmarianateixeira · Answer 2 · 2022-02-03T15:03:31+01:00

In this exercise we have to use the knowledge of internal energy, in this way we find that:

- 1.19 kJ

For this we have to organize the information given in the statement as:

Calculating internal energy as:

[tex]Qcal = Ccal *7.32 C\\Ccal = 3.50 kJ /7.32 C = 0.478 kJ/C\\Qcal = 0.478 kJ/C *2.49 C = 1.19 kJ\\Qreac = -Qcal = - 1.19 kJ\\ΔU = - 1.19 kJ[/tex]

See more about intrernal energy at brainly.com/question/1932868