Analysis Of Enthalpy
The specific heat, s, is known for most substances. Another equation needed is for calculating heat for objects with a fixed mass, like the calorimeter: q=C∆T. When these two equations are plugged into the first one, and it is expanded out, the resulting equation looks like this:
“C_cal (T_f-T_(i,cw) )=-(m_cw )(s_water )(T_f-T_(i,cw) )-(m_hw)(s_water)(T_f-T_(i,hw))” (French et al, 105).
Note that all of the substances reach the same final temperature. Temperature was measured during the reaction, allowing the amount of heat absorbed by the solution and the calorimeter to be measured. When added together, those values the enthalpy of the reaction. If the system loses no heat, this equation is true:
“q_rxn=-q_soln-q_cal” (French et al, 105).
This equation is equal to this one:
“q_rxn=-m_soln s∆T-C_cal ∆T” (French et al, 105).
For this reaction, s will be the specific heat of water because the experiment involves aqueous solutions. Also note that ∆T is the same. Once the heat of the reaction has been calculated, the enthalpy can be calculated by dividing the heat by the moles of limiting …show more content…
For reaction 1, which was between NaOH and HCl, the enthalpy of the reaction was -45.7 kJ/mol. For reaction 2, which was between NaOH and NH_4 Cl, the enthalpy of the reaction was -8.67 kJ/mol. For reaction 3, which was between HCl and NH_3, the enthalpy of the reaction was -46.2 kJ/mol. The enthalpy of reaction 3 determined by Hess’s Law and the results of reactions 1 and 2 was -37 kJ/mol. The percent error for the enthalpy of reaction 3 acquired from the in-lab procedure was 13%. The percent error for the enthalpy of reaction 3 acquired using Hess’s Law was 30%. One significant correlation between all of the enthalpies calculated is that all of them are negative, showing that heat was lost in all of the reactions, making them exothermic. As for the percent errors, there was less error when the data acquired was recorded by observing a performed reaction. Guessing the enthalpy using Hess’s Law and the results of the other reactions introduced more error into the calculation. One source of error could be that when the reactant was being poured into the calorimeter, leaving the lid open for too long let too much heat loose, which would affect the results. This would have affected reactions 1 and 2, and that increased percent error would affect the enthalpy of reaction 3 calculated using Hess’s Law. It affected the enthalpy calculated from the in-lab results less because the was only one reaction where that