Question

In a coffee-cup calorimeter, 100.0 mL of 1.76 M HNO3 and 100.0 mL of 1.22 M Ca(OH)2 are mixed. Both solutions were originally at 24.6 °C. The maximum temperature observed during the experiment is 36.4 °C. Calculate the enthalpy change for the neutralization reaction that occurs. Assume the mixture in the coffee cup has the same density and specific heat as pure water.

*This is from a problem sheet that's for extra practice. It gives me the answer of -112 kJ. I get to q solution = 200g X 4.184 X 11.8, which equals 9.87 kJ, so q rxn = -9.87 kJ but I don't know where to go from here or if this is even correct so far. I should also mention that for some reason, this class doesn't do that extra calculation to account for heat lost to the cup.

Answer 1

There is no any information about heat absorbed by cup. Hence we assume 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 = C x m x T

Where , C is a specific heat of solution ,

m is a mass of solution ,

T is a change in temperature of solution.

Here, Volume of solution = Volume of acid + Volume of base = 100 ml + 100 ml = 200 ml

Density of the solution is 1 g / ml . Hence,mass of solution = 200 g

Specific heat of solution = Specific heat of water = 4.184 J / g ⁰ C   

T = T final - T initial = 36.4 ⁰ C - 24.6 ⁰ C = 11.8 ⁰ C   

Hence, q solution = 200 g x 4.184 J / g ⁰ C x 11.8 ⁰ C   

= 9874.2 J

Therefore, q reaction = - q solution

= - 9874.2 J

= 9.874 kJ

Consider a reaction 2 HNO ₃ + Ca(OH) ₂   Ca(NO ₃ ) ₂ + 2 H₂O

From reaction , Stoichiometric ratio = No of moles of Acid / No of moles of Base = 2/1 = 2

No of moles of HNO ₃ = Molarity x volume of solution in L

= 1.76 mol / L x 0.1 L

= 0.176 mol

No of moles of Ca(OH) ₂ = 1.22 mol / L x 0.1 L

= 0.122 mol

From reaction , No of moles of Acid = No of moles of base x stoichiometric ratio

= 0.122 x 2

= 0.244

No of moles of HNO ₃ = 0.176 mol are less than required 0.244 mol. HNO ₃ is a limiting reactant.Hence , 0.176 mol acid is neutralised.

For the reaction of 0.176 mol  HNO ₃ , q = - 9.874 kJ. Conversion factor is - 9.874 kJ / 0.176 mol HNO ₃

Hence, H reaction = 1 mol HNO ₃ x - 9.874 kJ / 0.176 mol HNO ₃ = - 56.10 kJ

ANSWER : H reaction = - 56.10 kJ