Question

17. A common laboratory reaction is the neutralization of an acid with a base. When 50.0 mL of 0.500 M HCl at 25.0 °C is added to 50.00 mL of 0.500 M NaOH at 25.0 °C in a coffee cup calorimeter, the temperature of the mixture rises to 28.2 °C. What is the enthalpy of reaction per mole of acid? Assume the mixture has a specific heat capacity of 3.89 J/g·° C and a density of 1.09 g/mL while the cup has a heat capacity of 29.1 J/°C.

Answer 1

Given:

Volume of HCl = volume of NaCl = 50.0 mL

Molarity of HCl = molarity of NaCl = 0.500 M

Initial T = 25 ⁰ C

Final T of =28.2 ⁰ C

Solution:

We have to calculate the enthalpy of the reaction

The relation between enthalpy and qrxn is as follow.

Delta Hrxn = qrxn/n

qrxn is the heat given or taken by the reaction.

Whatever the heat given by the reaction will be taken by the solution. (reaction mixture)

So we write relation between qrxn and q sol

q.rxn = - qsol – qcal

we use following formula to find qsol

q = m x C x Delta T

m is mass in g , C is specific heat , Delta T is change in T.

we have to find mass of solution and moles of limiting reactant.

Mass of solution = volume in mL x density

= 100 mL x 1.09 g/mL =109.0 g

Calculation of moles of limiting reactant.

Since both reactants have 1 to 1 mol ratio and both have same amount of volume and same molarity.

We can use any of the reactant to find moles.

Moles of acid = 0.5 M x 0.050 L = 0.0050 mol

qsol = 109.0 g x 3.89 J/⁰C x (28.2 – 25.0)

= 1356.83 J

Now we calculated qcal

q cal = 29.1 J/deg C x (28.2-25.0 )

= 93.12 J

q rxn = - qsol-qcal = -1356.83 J-93.12 J

Delta H = -1356.83 -93.12 / 0.0050 mol

= -271459.52 J/mol

=- 271.5 kJ/mol

So the heat of reaction = -271.5 kJ /mol