1. The heat of neutralization is the heat of reaction when one mole of acid reac

Question

1. The heat of neutralization is the heat of reaction when one mole of acid reacts with one mole of base to create one mole of salt and one mole of water. Use the following data to determine the heat of neutralization (in kilojoules per mole) of HCl by NaOH

For this problem you may ssume that the densities of the solutions of HCl, NaCl, and NaOH are 1.000 g/mL.

Data:

Mass of 1.02 M aq. HCl used = 100.0 g

Mass of 1.02 M aq. NaOH used = 100.0 g

Temperature of solutions before mixing = 17.39 °C

Temperature of solutions immediately upon mixing = 24.42 °C

Heat capacity of the calorimeter = 43.39 Joules / °C

2. Use the proper graphing method (see Appendix E: Graphing Techniques in the lab manual), including metric graph paper, title, labelling of the axes, units and scaling, to plot a graph of temperature versus time of the following data:

Explanation / Answer

Mass of the solution = masss of NaOH+ mass of HCl= 100+100=200gm

temperature of solution before mixing= 17.39 deg.c, temperature of solution after mixing = 24.42 deg.c

Change in temperature= 24.42-17.39= 7.03 deg.c

Heat added to the solution = mass of solution* specific heat* change in temperature= 200*3.995*7.93 joules=5617 joules

heat lost to calorimeter= heat capacity of calorimeter* change in temperature =43.39*7.03 joules=306 joules

hence enthalpy change= heat added to the solution- heat lost to calorimeter= 5617-306= 5311 joules

mass of HCl used= 100 gm. moles of HCl= mass/molar mass= 100/36.5= 2.74

enthalpy change/moles of HCl= 5311/2.74 =1938 J/mole

the plot of temperature ( y axis) and time (x axis) is drawn and shown below.

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