You mix 125 mL of 0.154 M CsOH with 60.0 mL of 0.456 M HF in a coffee cup calorimeter, and the temperature of both solutions rises from 24.4°C before mixing, to 35.9°C after the reaction. CsOH (aq) + HF (aq) - -> CsF (aq) + H2O (l) What is the heat [q] for this reaction per mole of CsOH? Assume the densities of the solutions are all 1.00 g/mL and the specific heats of the solutions are 4.18 J / (g·K).

the heat of reaction per mole of CsOH is Δh = 461 kJ/mol

Explanation:

Since the number of moles present in the mixture is

n CsOH = 0.154 M * 0.125 L = 0.01925 moles

n HF = 0.456 M * 0.060 L = 0.02736 moles

then since a coffee cup calorimeter is a constant pressure calorimeter, from the first law of thermodynamics applied to enthalpy:

ΔH = Q - ∫VdP, since P=constant →dP=0 → ∫VdP=0, then

ΔH = Q

where

ΔH = enthalpy change (heat of reaction)

Q = m*cp * (T final - T initial)

where

m = mass of the solution

cp = specific heats

T final = final temperature

T initial = initial temperature

then since the density ρ is the same for both components

m = ρ * V total, where V total = volume of the solution

therefore

ΔH = m*cp * (T final - T initial) = ρ * V total * cp * (T final - T initial)

replacing values

ΔH = ρ * V total * cp * (T final - T initial) = 1.00 g/mL * (125 ml + 60 ml) * 4.18 J / (g·K) * (35.9°C - 24.4°C) = 8892.95 J

the heat of reaction per mole of CsOH is

Δh = ΔH / n CsOH = 8892.95 J / 0.01925 moles = 461971 J/mol = 461 kJ/mol

Δh = 461 kJ/mol