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SUSB-053

1. CALIBRATION - DETERMINATION OF THE CALORIMETRIC CONSTANT

How do we calculate T when two samples of water of known mass and temperature are mixed?

E.g., suppose

60 g of hot water at 50oC, are added to 30 g of cold water at 25oC in a calorimeter

Ideally,
 
heat gain by cold water   
=
-  heat lost by hot water

Since heat gained or absorbed is positive while heat lost is negative, we must put a negative sign on the heat lost for the equality to hold.

 
qcold    =      mc Cc ΔTc  
=
- qhot          =      - mh Ch ΔTh
 
30 g * 4.18 J/g * ( Tmix - 25 ) deg 
=
- 60 g * 4.18 J/g * (Tmix - 50 )
 
30 * ( Tmix - 25
=
- 60 * ( Tmix - 50 )
 
 Tmix 
=
( 60 * 50 + 30 * 25 ) / ( 30 + 60 )
   
=
42oC
 

Note: If the volumes of hot and cold water are identical (as in the exercise), the final temperature, Tmix, is simply the average of the temperatures of the hot and cold water. For arbitrary volumes of hot and cold water, the final temperature is the mass weightd average of the hot and cold water.

 

For a real system, the actual  Tmix will differ from this value because of some heat loss to the container - i.e., to the calorimeter.

 

Assume the calorimeter experiences the same temperature change as the cold water. (The cold water was originally in, and at the same temperature as, the calorimeter.) This is determined experimentally by measuring the temperatures of the relevant water samples before and after mixing. (Example of such a determination.) We also assume that the final calorimeter temperature is the same as that of the final solution.

The heat lost by the hot water now goes to increasing temperature of both the cold water and the calorimeter.

Now we have

 
- qhot 
=
 qcold + qcal
  or    
 
qcal 
=
 - qhot - qcold
 

Suppose, for example, that the actual (measured) final temperature, Tmix, of the same above mixture is 40oC

- qhot  = - 60 g * 4.18 J/g * ( Tmix - 50 ) = 2510 J

  qcold =   30 g * 4.18 J/g * ( Tmix - 25 ) = 1880 J

Note that since some heat must be transferred to the calorimeter, the magnitude of the heat lost by the hot water must be greater than the magnitude of the heat gained by the cold water.

qcal = -qhot - qcold = 2510 - 1880 = 630 J

ΔT for the calorimeter = ΔT for the cold water = 15 Co

It behaves like an object with a "total specific heat" of

Ccal = qcal / ΔT = 630 / 15 = 42 J / Co

 

2. DETERMINATION OF ΔHneu FOR HCl

It is essential to keep track of which quantitites are actually measured in this exercise and which ones are given.

Suppose we mix samples of measured volumes and measured temperatures and given concentrations of HCl and NaOH

VHCl = 50.0 mL
VNaOH = 55.0 mL
THCl = 24.0oC
TNaOH = 24.2oC
MHCl = 1.20 M
MNaOH = 1.30M

We measure temperature change at t = 0 graphically. (Here is the graph.) The result is that ΔT = 4.75Co

We need to know how much of the essential reactant (HCl) has been consumed. That requires determining the limiting reagent.

Using the measured volumes and given concentrations of HCl and NaOH, we can calculate the number of moles of each reactant.

 

 
nHCl       
= VHCl   * MHCl   = 0.0600 mol
 
nNaOH    
= VNaOH * MNaOH = 0.0715 mol

 

This makes HCl the limiting reagent. Only 0.0600 mol HCl can be neutralized.

Using the given heat capacity, given densities and measured volumes of the solutions and the measured temperature change, we can analyze the heat exchange, q, and correct for the heat loss to the calorimeter using the Calorimetric constant, Ccal, determined in Part 1.

Heat Capacity = 3.97 J/g
Density = 1.02 g/mL
Ccal = 42.0 J/C
o

Tot Vol of Sol = 50.0 + 55.0 = 105.0 mL

(We ignore a smalll additional volume of water (0.060 mol = 1.1 g) which is generated by the neutralization reaction.)

From the given density, we can calculate the total weight of this volume of solution

Tot Wt Sol = 1.02 * 105.0 = 107 g

From the given heat capacity of the solution and the measured temperature change, we can calculate the amout of heat absorbed by the solution.

qSOL = 3.97 J/g Co * 107 g * 4.75 Co = 2020 J (actually, only 3 significant figures are justified)

But, some of the liberated heat went to raising the temperature of the calorimeter. Using the calorimetric constant and the measured temperature change, we can calculate this additional amount of heat absorbed by the calorimeter.

qCAL = 42.0 J/Co * 4.75Co = 200 J

The total number of Joules liberated by the reaction is the sum of the amount absorbed by the solution and by the calorimeter, -(qSOL + qCAL),

                                                                                                                qRXN = - (2020 J + 200 J) = -2220 J                       (liberated heat is negative)   

But the reaction involved the neutralization of 0.0600 mol of HCl. therefore, the molar heat of neutralization was

qRXN = -2200 J / 0.0600 mol = -3.67 X 104 J/mol

But a positive change in enthalpy for a process represents heat gained (q) by the system. The reaction liberated heat rather than gaining it. The heat, and therefore, the enthalpy of the neutralization reaction is the negative of the above amount, or

ΔHneut for HCl = - 3.67 X 104 J/mol

The analysis for the neutralization of the weak acid is identical to that above.

 

The remainder of the calculations involve the application of Hess' Law to obtain the enthalpy of dissociation.

 

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Last Update: RFS 2016-10-14