Question

A volume of 500.0 mL of 0.220 M HCl(aq) was added to a high quality constant-pressure calorimeter containing 500.0 mL of 0.200 M NaOH(aq). Both solutions have a density of 1.000 g mL-1 and a specific heat of 4.184 J g‑1 oC-1. The temperature of the entire system rose from 25.60 °C to 26.70 °C. Calculate the heat of reaction, in kJ, per mole of NaOH(aq).

Answer 1

Heat of the reaction per mole of NaOH = 46.02 kJ/mol

Step-by-step explanation:

The reaction between HCl (strong acid) and NaOH(strong base) is a neutralization reaction which yields a salt 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 = 500.0 ml

Volume of NaOH = 500.0 ml

Density of HCl and NaOH = 1.000 g/ml

`Mass = Density*Volume`

`Mass(HCl) = Mass(NaOH) = 1.000g/ml*500.0ml = 500.0 g`

Total mass of the solutions, m = 500.0 +500.0 = 1000.0 g

c = 4.184 J/g/c

T1 = 25.6 C

T2 = 26.70 C

Substituting appropriate values in equation (1) gives:

`q = 1000.0 g*4.184 J/gC *(26.7-25.6)C = 4602.4 J=4.602kJ`

Now, the number of moles of NaOH is:

`Moles(NaOH)=Molarity(NaOH)*Volume(NaOH)= 0.200moles/L*0.500L = 0.100moles`

Heat of reaction/mole NaOH is:

`=(q)/(moles(NaOH))=(4.602kJ)/(0.100moles)=46.02kJ/mol`