Thanks much Begin a new page in your laboratory notebook and answer the followin
ID: 1067748 • Letter: T
Question
Thanks much
Begin a new page in your laboratory notebook and answer the following questions. Explain your reasoning and write legibly. It is not necessary to copy the questions into your notebook, but clearly identify each question by number. How well did your experimental value for the enthalpy of solution compare with the calculated value using data from the Handbook of Chemistry & Physics? How would you explain the difference in these values? Address each of the following: Write a balanced equation for the formation of one mole of NaCl(s) from its elements. delta H degree f= -411.1 kJ/mol. Write a balanced equation for the formation of one mole of NaCl(aq) from its elements. delta H degree f = -407.3 kJ/mol. Write a balanced equation for the conversion of solid NaCl(s) to aqueous NaCl(aq). Using Hess's Law, and the previous three equations, calculate delta H degree_soln, in kJ/mol, when NaCl(s) dissolves in water. Is this process highly or slightly exothermic or endothermic? For the two neutralization reactions studied, was the amount of heat released the same or different? Did you expect them to be the same or different? Explain your rationale. The dissociation of HC_2H_3O_2 is represented below in Equation 1. This process involves a small energy change that is very difficult to measure: HC_2H_3O_2(aq) + H_2O(l) rightarrow H_3O^+(aq) + C_2H_3O_2^-(aq) equation 1 In lab you determined the enthalpy values for the following reactions: H_3O^+(aq) + OH^-(aq) rightarrow 2H_2O(l) Equation 2 HC_2H_2O_2(aq) + OH^-(aq) rightarrow C_2H_33O_2^- (aq) + H_2O(l) Equation 3 Use Hess's Law and your experimental results from Part III (Equations 2 and 3) to calculate bW for reaction (1).Explanation / Answer
(a)
Na (s) + 1/2 Cl2 (g) ------------> NaCl (s), Delta H0f = - 411.1 kJ/mol
(b)
Na (s) + 1/2 Cl2 (g) ----------> NaCl (aq.) , DeltaH0dil = - 407.3 kJ/mol
(c) (b) - (a)
NaCl (s) ----------> NaCl (aq.) ,
(d) DeltaH0sol = - 407.3 - ( - 411.1 ) = + 3.8 kJ/mol
Since enthalpy change is positive, it is endothermic.
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