Calorimetry heat of reaction

Question

A student adds 75.0 mL of a 1.80 M HCl to 75.0 mL of 1.80 M NaOH inside a calorimeter (C=77.0 J/C) and immediately places the lid on top. The solution temperature rises from 20.5C to 24.7C. The specific heat of water in 4.184J/gC and the density of water is 1.00 g/mL. Assume the density of the acid and base solutions are also 1.00 g/mL.

a. Calculate the heat of reaction

b. Calculate the molar enthalpy for the reaction

Answer 1

The temperature rises, so this is an exothermic reaction. Heat of reaction is the heat generated by the reaction taking place, under the given conditions. Some of the heat goes into the calorimeter, and some goes to the surroundings.

The amount of heat absorbed by the calorimeter is q = C_cal∆T

The amount of heat going to the surroundings is q = mC∆T

Adding these together to get the total heat of the reaction, we have

q = mC∆T + C_cal∆T

q = heat = ?

m = mass = 75.0 ml + 75.0 ml = 150.0 ml x 1 g / ml = 150.0 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 24.7º - 20.5º = 4.2º

Solving for q, we have ...

q = (150 g)(4.184 J/gº)(4.2º) + (77.0 J/º)(4.2º) = 2636 J + 323 J

q = heat of reaction = 2959 J (would be negative since it is an exothermic reaction)

Now, to find the MOLAR enthalpy for the reaction, we must find the moles of reactants and products for this particular reaction. Let's look at the balanced equation:

HCl + NaOH ==> NaCl + H2O ... balanced equation (NOTE: all reactants and products are in a 1:1 ratio)

Calculate moles:

moles HCl = 75.0 ml x 1 L / 1000 ml x 1.80 mol / L = 0.135 mols

mols NaOH = 75.0 ml x 1 L / 1000 ml x 1.80 mol / L = 0.135 mols

Molar enthalpy of reaction = ∆H_rxn = 2959 J / 0.135 mols

∆H_rxn = 21,919 J / mol = 21.9 kJ / mole (would be negative since it is an exothermic reaction)