In PART(B), They mixed 150mL each of 0.65M HCl and 0.65M NaOH, both are at tempe

Question

In PART(B), They mixed 150mL each of 0.65M HCl and 0.65M NaOH, both are at temperature of 18.1 degree C in a calorimeter equilibrated to the same temperature. After following the temperature change for 10 minutes, they found the final temperature of mixing to be 22.3 degree C. {Using the C_cal from of PART A, and assuming that the density of the of the two solutions is 1.00 g/mL. and their specific heat is 4.184 J/g- degree C}, Calculate the heat absorbed by the two solutions (qsol) Calculate the heat absorbed by the calorimeter, (qcal)? What is the heat change for the neutralization reaction (qrxn)?

Explanation / Answer

a)

Heat absorbed

Q = m*C*(Tf-Ti)

m = 150 + 150 mL = 300 mL, assume 1 g/mL = 300 g

Q = 300*4.184*(22.3-18.1) = 5271.84 J

b)

heat by calorimeter...

impossible with no Ccal...

Q = Ccal*(Tf-Ti)

Q = Ccal(22.3-18.1)

c)

heat change for neutralization

assume Qcal = 0 approx

so

Qsolutions = -Qreaction

Qreaction = -5271.84 / (mol)

mol of HCl = MV = 0.15*0.65

mol of HCl = 0.0975

Qreaction =  -5271.84 / (0.0975) = -54070.1538 J = -54.07 kJ/mol which is pretty near to the real value of -56 kJ

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