I did a chem lab where the enthalpy for various solutions mixed together in a ca

Question

I did a chem lab where the enthalpy for various solutions mixed together in a calorimoter was calculated. Here are my results- HCL + NaOH= NaCl +H2O enthalpy= -71.6 Kj/mol

CH3COOH + NaOH= CH3COOHNa + H2O enthalpy= -61.368 Kj/mol

CH3COOH + NH3= NH4CH3COO enthalpy= -56.254

use this data to caluculate the enthalpy for HCL + NH3= NH4Cl. I got the enthalpy to equal -66.486.

1. Using the literature values provided below calculate the change in enthalpy for this reaction (HCL + NH3=NH4Cl)

HCL= -167.159 KJ/mol

NH3= -80.835 KJ/mol

NH4Cl= -299.66 KJ/mol

2. Compare your calculated value witht he value you calculated using your experimentally generated data. Are the two values in cole agreement? If no, why?

3. The heat eveolved for the neutralization of HCL with NaOH was greater than that evolved for the neutralization of CH3COOH with NaOH. Why? Should this have been expected?

4. How would your data have been affected if you had failed to dry your calirometer between experiment?

Explanation / Answer

1) HCL + NH3= NH4Cl

delta H rxn = delta H products - delta H reactants

=> delta H rxn = (-299.66) - (-80.835 - 167.159) = - 51.666 kJ / mol (From Literature)

2)

Experimental delta H

HCL + NaOH= NaCl +H2O enthalpy= -71.6 Kj/mol

CH3COOH + NH3= NH4CH3COO enthalpy= -56.254

CH3COONa + H2O = CH3COOH + NaOH enthalpy= 61.368 Kj/mol

Adding the three reactions we get,

HCL + NH3= NH4Cl

delta H = -71.6 - 56.254 + 61.368 = -66.486 kJ / mol (Experimental)

Enthalpy (Theoretical) = -51.666 kJ / mol

Enthalpy (Experimental) = -66.486 kJ / mol

The difference in the values are due to the non ideal experimental conditions. Heat is lost to he environment during the reaction which in not accounted for. Hence the difference in values

3) HCl and NaOH are strong Acid and Base respectively. Hence they completely neutralize each other. On the other hand CH3COOH is a weak acid. Hence complete neutralization does not take place. Hence heat evolved for the neutralization of HCL with NaOH was greater than that evolved for the neutralization of CH3COOH with NaOH

4) If you failed to dry your calorimeter, the remaining water in the calorimeter will absorb heat not accounted for in the calculations. The heat evolved values will decrease.

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