A student was assigned the task of determining the enthalpy changes for the reac

Question

A student was assigned the task of determining the enthalpy changes for the reaction between solid MgO and aqueous HCl represented by the net-ionic equation above. The student uses a polystyrene cup calorimeter and performs four trials. Data for each trial are shown in the table below. (a) Which is the limiting reactant in all four trials. HCl or MgO? Justify your answer. (b) The data in one of the trials is inconsistent with the data in the other three trials. identify the trial with inconsistent data and draw a line through the data from that trial in the table above. Explain how you identified the inconsistent data. For parts(c) and (d), use the data from one of the other three trials (i.e., not form the trial you identified in part (b) above). Assume the calorimeter has a negligible heat capacity and that the specific heat of the contents of the calorimeter is 4.18 J/(g middot C degree). Assume that the density of the HCl(aq) is 1. g/mL. (c) Calculate the magnitude of q, the thermal energy change, when the MgO was added to the 1.0 M HCl(aq). Include units with your answer. (d) Determine the student's experimental value of Delta H degree for the reaction between MgO and HCl in units of kJ/mol. (e) Enthalpies of formation for substances involved in the reaction are shown in the table below. Using the information in the table, determine the accepted value of Delta H degree for the reaction between MgO(s) and HCl(aq) (f) The accepted value and the experimental value of not agree. If the calorimeter leaked heat energy to the environment, would it help account for the discrepancy between the values? Explain.

Explanation / Answer

a)

mol of HCL = MV = 0.1*100 = 10 mmol of HCl

mol of MgO = mass/MW = 0.25/40.3044 = 0.00620 mol = 0.00620*10^3 = 6.2

mol of MgO = mass/MW = 0.50/40.3044 = 0.01241 mol = 0.01241 *10^3 = 12.4

ratio is:

1 mol of MgO per 2 mol of H+ so..

clearly, H+ ions are always limiting

b)

Calculate Q for each

Q1 = m*C*(Tf-Ti) = (100)*4.184*(26.5-25.5) = 418.4 J per 10 mmol of HCl

Q2 = m*C*(Tf-Ti) = (100)*4.184*(29.1-25) = 1715.44 J per 10 mmol of HCl

Q3 = m*C*(Tf-Ti) = (100)*4.184*(28.1-26) = 878.64 J per 10 mmol of HCl

Q4 = m*C*(Tf-Ti) = (100)*4.184*(28.1-24.1) = 1673.6 J per 10 mmol of HCl

The data which is not fitting is the firs tone, since the dT is too small compared to the others

c)

Q when MgO is added to HCl

Q2 = m*C*(Tf-Ti) = (100)*4.184*(29.1-25) = 1715.44 J per 10 mmol of HCl

Q3 = m*C*(Tf-Ti) = (100)*4.184*(28.1-26) = 878.64 J per 10 mmol of HCl

Q4 = m*C*(Tf-Ti) = (100)*4.184*(28.1-24.1) = 1673.6 J per 10 mmol of HCl

average = (1715.44 +878.64 +1673.6 )/3 = 1422.56 J

d

Experimentl

mol of HCL = 10 mmol = 10*10^-3

HRxn = Q/n = 1422.56 / (10^-2) = 142256 J/mol of HCl = 142.26 kJ/mol

e

HRxn= Hprod - Hreact = (Mg+2) + H2O - (MgO + 2H+) = -467+-286 -(0+-602) = -151 kJ/mol

f.

The heat is not that large, meaning it is missing some energy, i.e the leak

so it could, but we would have to know the magnitud to get a correct value

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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