Question

Q2: When 500.0 mL of 2.00 M Ba(NO3)2 solution at 25.0 °C is mixed with 500.0 mL of 2.00 M Na2SO4 solution at 22.0°C in a coffee-cup calorimeter, the white solid BaSO4 forms, and the temperature increases to 28.2°C. Assume that the calorimeter is insulated and has negligible heat capacity, the specific heat capacity of the solution is 4.184 J/g°C, and the density of the final solution is 1.0 g/mL. Answer the questions below using the Balanced Equation: Ba(NO3)2 (aq)+ Na2SO4 > BaSO4 (s) + 2NaNO3 (aq). Answer to correct amount of Significant Figures

Determine the heat of the reaction in J

Answer 1

We can assume that , no heat is transferred to calorimeter.  Therefore, heat lost by reaction is equal to heat absorbed by the reaction. Hence we can write, q reaction = - q solution

We know that, q solution = m C T

Where, m is a mass of solution , C is a specific heat of solution, T is a change in temperature of solution.

To find out q solution , we must know m , C and T.

After mixing barium nitrate with sodium sulfate, volume of solution will be the sum of volumes of each solutions.

Volume of solution = 500.0 ml + 500.0 ml = 1000.0 ml

We have formula, density = Mass / volume

Mass of solution = density volume = 1.0 g/ ml 1000.0 ml = 1.0 10 ³ g

T = T final - T initial = 28.2 ⁰ C - 22.0 ⁰ C = 6.2 ⁰ C

Substituting m = 1.0 10 ³ g , C = 4.184 J / g ⁰ C and T = 6.2 ⁰ C in q solution = m C T , we get

q solution = 1.0 10 ³ g 4.184 J / g ⁰ C 6.2 ⁰ C

q solution = 25940.8 J

q solution = 25.94 k J

We have, q reaction = - q solution

q reaction = - 25.94 k J

Now, Consider reaction Ba( NO₃ ) ₂ (aq)  + Na₂SO₄ (aq) BaSO ₄ (s) + 2 NaCl (aq)

From above reaction, 1 mol Ba( NO₃ ) ₂   1 mol Na₂SO₄ 1 mol BaSO ₄

We can calculate no. of moles by using formula , No of moles of = Molarity   volume of solution in L

No. of moles of Ba( NO₃ ) ₂ = 2.00 mol / L 0.5000 L = 1.00 mol

No. of moles of Ba( NO₃ ) ₂ = No. of moles of Na₂SO₄ =No. of moles of BaSO ₄ = 1.00 mol

For the reaction of 1.00 mol Ba( NO₃ ) ₂  , q = -25.94 kJ.

ANSWER: H reaction = - 26 k J