Bonus 1. Calculate the molar enthalpy of neutralization (delta H) in kj/mol of t

Question

Bonus 1. Calculate the molar enthalpy of neutralization (delta H) in kj/mol of the exothermal reaction between an aqueous acid and an aqueous base (assume 1:1 stoichiometry) given the following information: Temperature change = 6.23 degree C 50.0 mL of 2.0 M acid added to 50.0 mL of 2.0 M base Heat capacity of the calorimeter (C _= 6.50 J/degree C Specific heat for water = 4.184 J/E degree C You must show all of your work report your answer to the proper number of significant figures, and the appropriate sign.

Explanation / Answer

dT = 6.23 °C

Total V = 50+50 = 100 mL

Asusme Density of solution = 1 g/mL

mass of solution = 100*1 = 100 g

mol of acid = MV = 50*2 = 100 mmol = 0.1 mol

Qsolution = m*C*dT

Qcalorimter = C*dT

Qtotal = Qsolution + Qcalorimter

Qtotal = Qreaction

Qtotal = 100*4.184*6.23 + 6.5*6.23 = 2647.127 J

so

Qreaction = -2647.127 J

then

HRxn = Qrxn/mol = -2647.127 / (0.1) =  -26471.27 J/mol

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---