When 10.0 mL of a 2.0 M acetic acid solution at 22 2 degree C is reacted with 10

Question

When 10.0 mL of a 2.0 M acetic acid solution at 22 2 degree C is reacted with 10.0 mL of a 2.1 M NaOH solution at 22.8 degree C in a coffee cup calorimeter, the resulting temperature of the solution was determined to be 27.9 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 .184J/g degree C) The calorimeter constant was 102 J/degree C. Select one: -450 J -55 J -510 J -400 J 2600 J

Explanation / Answer

HRxn = -Q/n

n = MV = 10 mL * 2 M = 20 mmol of acetic acid

ratio is 1:1 acid:base so,

20 mmol of acid -->20*10^-3 = 0.02 mol

then..

Q = m*C*(Tf_Ti) = 20*4.184*(27.9-22.2) = 476.976 J

for calorimeter

Q = C*DT = 10.2 * (27.9-22.2) = 58.14 J

total Q = 58.14+476.976 = 535.116 J

so..

HRxn = -535.116/(0.02) = 26755.8 J

nearest answer is E

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