A coffee cup calorimeter contains 100.0 mL of 1.50 M Ba(NO3)2 at 25.0o C. A student pours 100.0 mL of 1.50 M Na2SO4 at 25.0o C into the calorimeter. A precipitate forms and the temperature rises to 29.7o C. Assume that no heat was lost to the surroundings, the volumes were additive, the specific heat capacity of the solution was 4.184 J/gK, and the density of the solution was 1.00 g/mL. A. This reaction could be represented as a complete ionic or a net ionic equation. Write the balanced chemical equation that should be used in association with the Delta Hrxn value for this reaction.
B. Find the amount of heat that was lost or gained by the solution in the calorimeter.
C. Find the heat of reaction, Delta Hrxn.
D. Is the reaction endothermic or exothermic?

Ba(NO₃)₂  +  Na₂SO₄   →  BaSO₄ + 2NaNO₃

q = 3.93 kJ

ΔH_{rxn} =  26.2 kJ/mole

Exothermic

Explanation:

Given that:

The Volume of Ba (NO₃)₂ = 100.0 mL

The molarity of Ba (NO₃)₂ = 1.50 M

The volume of Na₂SO₄ = 100.0 mL

The molarity of Na₂SO₄ = 1.50 M

From the above, the total volume of solution = 200.0 mL

Density of solution = 1 g/mL

The balanced equation can be written as:

Ba(NO₃)₂  +  Na₂SO₄   →  BaSO₄ + 2NaNO₃

We know that:

Density = mass/volume

Thus;

the mass of the solution = density × volume

= 200 × 1

= 200 g

Temperature change ΔT = T₂ - T₁

ΔT = (29.7 - 25)° C

ΔT = 4.7° C

Heat capacity c = 4.184 J/gK

∴

The amount of heat lost or gained q = m× c×ΔT

q = 200 × 4.184 × 4.7

q = 3932.96 J

q = 3.93 kJ

From the above information, we can determine the number of moles of Ba(NO₃)₂ and  Na₂SO₄

For Ba(NO₃)₂

number of moles = 0.10 × 1.50

= 0.15

For Na₂SO₄

number of moles = 0.10 × 1.50

= 0.15

∴

ΔH_{rxn} = q/number of moles

ΔH_{rxn} = 3.93/0.15

ΔH_{rxn} =  26.2 kJ/mole

From the reaction, we will notice that the temperature increased. This significantly implies that heat is evolved from the reaction. As such, the reaction is exothermic.

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