When 10.0 mL of a 2.0 M acetic acid solution at 20.6 degree C is reached with 10

Question

When 10.0 mL of a 2.0 M acetic acid solution at 20.6 degree C is reached with 10.0 mL of a 2.1 M NaOH solution at 21.0 degree C in a coffee cup calorimeter, the resulting temperature of the solution was determined to be 28.6 degree C. Calculate the heat of reaction for this acid base neutralization. Assume the density and heat capacity of the solution to be the same as water (d = 1.00 g/mL and c_p = 4.184 g degree C). The calorimeter constant was 2.7 j/degree C Select one: a -650 j b. -670 j c. -2500 j d. - 630 j e. -21 j

Explanation / Answer

1st calculate the heat absorbed by solution
Qsolution = Q(acetic acid) + Q (NaOH) + Q (calorimeter)
= 10.0 g * 4.184 J/goC*(28.6-20.6) oC + 10.0 g * 4.184 J/goC*(28.6-21) oC + 2.7 J/oC*(28.6-21) oC
= 334.72 J + 317.98 + 20.52
= 673.22 J
This heat is released by reaction

so, heat of reaction = -673.22 J
Answer: b

Since this is closest

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