Question

When acid and water are mixed, heat is released which can cause the solution to boil and splatter. Calculate the final temperature of the solution when 10.00 mL of 2.50 M HCN at 25°C is added to 200 mL of water at 25°C. Assume that: HCN releases 50 kJ/mol when dissolved, the solution has a density of 1.01 g/mL and a specific heat capacity of 4.18 J/g°C, and that the container has a negligible heat capacity.

(1) 2.3°C (2) 23.5°C (3) 25.0°C (4) 26.4°C (5) 40.3 °C

Answer 1

We are given a solute + solvent. In this case is an ACID + WATER respectively.

They will break solute-solute and solvent-solvent interacitons in order to yield either a positive or negative value of solvent-solute interaction energy. This is called enthalpy/heat of solution/solvation.

For HA + H2O = H3O+ A- in solution

H-Sol = -Qevolved by solvation / mol of solute

-Qevolved = Qsolution

We can relate the solvation with the solution's heat

we can do this via

Qsolution = m-solution * C-solution * (Tfinal - Tiniail)

and we know that, mass of solution

mass of solution = Density of solution * Total Volume of solution

C-solution = assume 4.18 J/gC (similar to water)

substitute this data

Now; the counterpar of the equation:

calculate moles of solute

mol = mass/MW or mol = M*V

mol = MV = (2.5 M)(0.01 L) = 0.025 mol of HCN

the equation so far:

H-Sol = -Qevolved by solvation / mol of solute

H-sol = -D*V*C* (Tfinal - Tiniail)/ mol of solute

50*10^3 = -(1.01)(200+10)*(4.18)(Tf-25) / 0.025

solve for Tf

(50*10^3)*0.025 / ((1.01)(200+10)*(4.18)) = Tf-25

1.40991 = Tf- 25

Tf = 25+1.40991 C

Tf = 26.40991 C

choose (4) 26.40991