Hello I need questions 9-14 answered please. Use the theoretical C p value (more

Question

Hello I need questions 9-14 answered please.

Use the theoretical Cp value (more precisely determined value) from Experiment 1 for your calculations in Experiment 2.

8. Record the following for each of the three trials. Be sure to record all data and calculated results with the correct number of significant figures.

Trial 1

a

Mass of the empty calorimeter (g):

18.6

b

Initial temperature of the HCl solution in calorimeter (

a

Mass of the empty calorimeter (g):

18.6

b

Initial temperature of the HCl solution in calorimeter (

Explanation / Answer

The relation of heat released q = m x C x delta T

where,

m = mass in g

C = specific heat in J/g.C

delta T = change in temperature in C

9. Now we calculate qwater for Trial 1,

qwater = 50 x 4.184 x 13

            = 2719.6 J

Similar calculations for Trial 2 and Trial 3 for qwater is done, we get the final values as,

Trial 1           qwater = 2719.6 J

Trial 2           qwater = 4566.8 J

Trial 3           qwater = 6379.5 J

10. Now calculate qcalorimeter or qcal by relation,

qcal = Ccal x delta T

For trial 1, we have

Ccal = 21.3 J/C

delta T = 13 C

Feeding all th values we get,

qcal = 21.3 x 13

        = 276.9 J

Similar calculations for Trial 2 and Trial 3 was done, overall the results are,

Trial 1           qcal = 276.9 J

Trial 2           qcal = 462.2 J

Trial 3           qcal = 645.4 J

11. To calculate qreaction we will use the relation,

qreaction = qcal + qwater

For Trial 1,

qreaction = 2719.6 + 276.9

                = 2996.5 J

Similar calculations for Trial 2 and Trial 3 was done, the results are,

Trial 1           qreaction = 2996.5 J

Trial 2           qreaction = 5029.0 J

Trial 3           qreaction = 7024.9 J

12. Enthalpy change delta H for the reaction

At constant pressure delta H = -qwater

The enthalpy change in kJ for each Trial run will be,

Trial 1           delta H = -2.72 kJ

Trial 2           delta H = -4.57 kJ

Trial 3           delta H = -6.38 kJ

For Trial 1, enthalpy change per mole Mg will be ,

moles of Mg 0.685 mol

delta H = -2.72 / 0.685

             = -3.97 kJ/mole of Mg

Similar calculation done for Trial 2 and Trial 3 gave result (delta H / mole of Mg),

Trial 1           delta H = -3.97 kJ

Trial 2           delta H = -6.63 kJ

Trial 3           delta H = -9.25 kJ

average enthalpy = -6.62 kJ/mole of Mg

14. The amount of heat released when 3.0 mol of Mg is reacted with excess HCl is,

= -6.62 x 3.0

= -19.86 kJ

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.