Question

A common laboratory reaction is the neutralization of an acid with a base. When 32.8 mL of 0.500 M HCl at 25.0°C is added to 66.3 mL of 0.500 M NaOH at 25.0°C in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°C. What is the heat of reaction per mole of NaCl (in kJ/mol)? Assume the mixture has a specific heat capacity of 4.18 J/(g·K) and that the densities of the reactant solutions are both 1.07 g/mL.

Answer 1

Step-by-step explanation:

The given data is as follows.

Densities of both the reactant solutions = 1.07 g/mL.

Specific heat = 4.18
`J/g^(o)C`

Volume of HCl = 32.8 ml

Volume of NaOH = 66.3 ml

The reaction will be as follows.

`HCl + NaOH \rightarrow NaCl + H_(2)O`

Since, density is mass divided by volume. Hence, calculate the mass of HCl as follows.

Mass of HCl = Density × Volume of HCl

=
`1.07 * 32.8 ml`

= 35.096 g

Similarly, mass of NaOH will be calculated as follows.

Mass of NaOH = Density × Volume of NaOH

=
`1.07 g/ml * 66.3 ml`

= 70.941 g

Therefore, total mass will be as follows.

Total mass = Mass of HCl + Mass of NaOH

= 35.096 g + 70.941 g

= 106.037 g

Change in temperature will be calculated as follows.

`\Delta T` =
`(28.2 - 25)^(o)C`

=
`3.2^(o)C`

Therefore, calculate the heat of reaction as follows.

q =
`mC \Delta T`

=
`106.037 g * 4.18 J/g^(o)C * 3.2^(o)C`

= 1418.35 J

or, = 1.418 kJ (as 1 kJ = 1000 J)

No. of moles of HCl = Molarity × Volume

= 0.5 M × 32.8 ml

= 16.4 mol

No. of moles of NaOH = Molarity × Volume

= 0.5 M × 66.3 ml

= 33.15 mol

So, HCl is the limiting reagent and heat of reaction produces by per mole of NaCl will be calculated as follows.

Heat released for 1 mole of NaCl =
`(1.418 kJ)/(16.4 mol)`

= 0.0864 kJ/mol

Thus, we can conclude that the heat of reaction per mole of NaCl is 0.0864 kJ/mol.