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Question

A coffee-cup calorimeter was calibrated using 10.0 mL of 0.100 M HCl and 2.00 mL of 0.300 M NaOH (ΔH^o_{neutralization}= -55.8 kJ/mol). The reaction mixture had 0.300^oC increase in temperature. Using the same calorimeter, 5.00 mL of 0.250 M HNO₃ was made to react with 7.00 mL of 0.100 M NaOH. A temperature rise of 0.320^o C was observed. What is the ΔH^o_reaction of this reaction?

Select one:

a.

~ –51.0 kJ/mol

b.

~ –35.8 kJ/mol

c.

~ +55.8 kJ/mol

d.

~ +51.0 kJ/mol

Answer 1

First, we need to find the C_cal, or calorimeter constant. We will then use this value in a subsequent calculation to find the ∆H for the reaction of interest. The C_cal is simply how much heat is absorbed by the calorimeter for each degree change in temperature, so it will have units of joules or kjoules per degree.

Finding C_cal:

HCl + NaOH ==> H₂O + NaCl

mols HCl = 0.0100 L x 0.100 mol/L = 0.00100 mols HCl

mols NaOH = 0.002 L x 0.300 mol/L = 0.0006 mols NaOH

NaOH is limiting so we use 0.0006 mols in our calculations

q = Ccal x ∆T

q = heat = 55.8 kJ/mol x 0.0006 mols = 0.03348 kJ = 33.48 J

C_cal = ?

∆T = 0.300º

C_cal = 33.48 J / 0.300º

C_cal = 111.6 J / degree

We will now use this to find the ∆H of the reaction of interest.

q = C_cal∆T

q = heat = ?

C_cal = 111.6 J/º

∆T = 0.320º

q = (111.6 J/º)(0.320º)

q = 35.71 J

HNO₃ + NaOH ==> H₂O + NaNO₃

mols HNO₃ = 0.005 L x 0.250 mol/L = 0.00125 mols HNO₃

mols NaOH = 0.007 L x 0.100 mol/L = 0.0007 mols NaOH

NaOH is limiting so we use 0.007 mols for our calculations

∆H_rxn = 35.71 J / 0.0007 mols

∆H_rxn = 51,014 J/mol = 51.0 kJ/mol (Answer d)

(note: ∆H_rxn has a positive value because the temperature increased meaning an endothermic reaction)