Question

A 1000 ml sample of 0.300 M NaOH is mixed with 100.0 ml of 0.300 M HCl in a coffee cup calorimeter. If both solutions are at 20.0 °c and the final temperature of the mixture was 22.0 °c Find the heat of neutralization, AHreu in kJ/mole. Assume no heat is lost to the surroundings, the density of all solutions is 1.00 g/mL, and C, of the mixture is 4.184 J/g·°C

Answer 1

ΔH(neu) = 30.68 kJ/mol

Step-by-step explanation:

The reaction between HCl (strong acid) and NaOH(strong base) is a neutralization reaction which yields sodium chloride NaCl and water

`HCl + NaOH \rightarrow NaCl + H2O`

The heat (q) of a reaction is given as:

`q = m*c*\Delta T=m*c*(T2-T1)-------(1)`

where m = mass of the system

c = specific heat

T1 and T2 are the initial and final temperatures

It is given that:

Volume of HCl = 1000 ml

Volume of NaOH = 100.0 ml

Density of HCl and NaOH = 1.000 g/ml

`Mass = Density*Volume`

`Mass(NaOH) = 1.000g/ml*1000 ml = 1000 g`

`Mass(HCl) = 1.000g/ml*100.0 ml = 100.0 g`

Total mass of the solutions, m = 1000 +100.0 = 1100.0 g

c = 4.184 J/g/c

T1 = 20.0C

T2 = 22.0 C

Substituting appropriate values in equation (1) gives:

`q = 1100.0 g*4.184 J/gC *(22.0-20.0)C = 9204.8 J=9.205kJ`

Now, the number of moles of NaOH is:

`Moles(NaOH)=Molarity(NaOH)*Volume(NaOH)= 0.300moles/L*1.000L = 0.300moles`

Enthalpy of neutralization is:

`=(q)/(moles(NaOH))=(9.205kJ)/(0.300moles)=30.68kJ/mol`