In: Chemistry
Use the following data to work through problems 4-6.
Initial Temp |
Final Temp |
Temp Change |
|
NaOH + HCL |
19.85 |
33.07 |
13.2 |
NaOH+ NH4Cl |
19.78 |
20.40 |
0.62 |
HCl + NH3 |
20.16 |
32.05 |
11.89 |
∆Hreaction = ∆Hsystem
% Error = Experimental – Accepted x 100
Accepted
In this case, the volume, concentration and density conditions for each of the solutions will be considered the same. Therefore, the values of mass and moles are determined for each reaction:
To calculate the enthalpy of reaction:
For the calculation for each mole, then:
And so for each reaction, getting:
To calculate the percentage deviation, it is necessary to calculate the standard formation enthalpy with theoretical values, using the equation:
According to the exercise, the values have already been calculated previously. In any case, here are the theoretical formation values (Kj/mol) for each compound:
NaOH | 469,4 |
HCl | 92,31 |
NaCl | 411 |
H2O | 285,84 |
NH3 | 67,20 |
NH4Cl | 315,4 |
For the reaction 1, so:
%
And so on:
As can be seen, the deviation percentages are very high. Remembering that it is perhaps necessary to perform a correction factor according to the ambient temperature, since the reaction enthalpies used are at a temperature of 25C. Or, verify the user values with the dice with the teacher and perform the same calculations again.