Acommon 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.

Explanation:

The given data is as follows.

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

  Specific heat = 4.18

  Volume of HCl = 32.8 ml

  Volume of NaOH = 66.3 ml

The reaction will be as follows.

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

             Mass of HCl = Density × Volume of HCl

                                   =

                                   = 35.096 g

Similarly, mass of NaOH will be calculated as follows.

            Mass of NaOH = Density × Volume of NaOH

                                     =

                                     = 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.

                =

                          =

Therefore, calculate the heat of reaction as follows.

                       q =

                           =

                           = 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 =

                                                          = 0.0864 kJ/mol

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

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j