2. When a neutralization reaction was carried out using 100.0 mL of o.7890M NH,

Question

2. When a neutralization reaction was carried out using 100.0 mL of o.7890M NH, water and 100.0 mLof 0.79MM acetic acid, AT was found to be476"C. The specific heat of the reaction mixture was 4.104 J g-1 K-1 . (1) Calculate Hneutzn for the reaction of NH3 and acetic acid. and its density was 1.03 g mL-1. The calorimeter constant was 3.36 JK- answer When it was calibrated, it was found that the thermometer read 0.50 C low. What effect would this thermometer reading have on the reported Hneutzn calculated in (2) above? (2) At the end of the experiment, it was discovered that the thermometer had not been calibrated.

Explanation / Answer

NH3 + CH3COOH = NH4CH3COO

Qsolution = m*C*(Tf-Ti) + CCal*dT

mass solution = (100+100)(1.03) = 206 g

Ccal = 3.36 J/K

Qsolution = (206*4.104)(4.76) + (3.36*4.76) = 4040.211 J absorbed

mol of base = MV = 100*0.789 = 78.9 mmol = 78.9 *10^-3

HRxn neutralization = -Qsolution / n = -4040.211 /( n) = ( -4040.211)/( 78.9 *10^-3) = -51206.73 J/mol

HRxn = -51.21 kJ/mol

Q2

if it reads 0.5 lower...

then,

there is no effect

since

dT = Tfinal - Tinitial = (Tfinal real + 0.5) - (Tinitial real + 0.5)

note that 0.5 - 0.5 = 0

cancel outs so no need to worry about it

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