-PRE-LAB: Show how; Reactions 1 and 2 add up to yield Reaction 3. Use that to sh
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-PRE-LAB: Show how; Reactions 1 and 2 add up to yield Reaction 3. Use that to show how the enthalpy for reactions 1 and 2 add up to give the enthalpy for reaction 3. Show the accepted values for the enthalpy of reactions 1-3 by summing the enthalpies of formation of reactants and products in these reactions. Your calculation will be easier if you use the net ionic equations. The following value is normally not listed on enthalpy tables because it is a reference value: AH?, H'(aq) - O kJ/mol. 1. 2. -OBJECTIVE: To determine the heat of reaction of three given reactions and use the results to confirm Hess's law. -INTRODUCTION: You will use the same calorimeter from Lab 10A to measure the heat released by three reactions. One of the reactions is the same as the combination of the other two reactions Therefore, according to Hess's law, the heat of reaction of the one reaction should be equal to the sum of the heats of reaction for the other two. This concept is sometimes referred to as the additivity of heats of reaction. The reactions we will use in this experiment are: 1. Solid sodium hydroxide dissolves in water to form an aqueous solution of ions NaOH(s) ? Na.(aq) + OH-(aq) ?Hi ? 2. Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride. NaOH(s) + H"(aq) + Cr(aq)-+ H2O(l) + Na"(aq) + Cr(aq) ??.-? Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride. 3. Na (a)+OH la) + H'la)+Cr(aa) Ho)+ Na (aa)+Crfaa) AH? Note that Reaction 3 is written as an ionic equation and not a net ionic so when you are cancelling species using Hess's law, do not cancel the spectator ions shown in Reaction 3.Explanation / Answer
1) In equation 1,NaOH (s) ----> Na+ (aq) + OH- (aq) ?H1
=> Na+(aq) + OH- (aq) ------> NaOH (s) - ?H1
Only the sign of enthalpy will change whereas the value is the same since both reactions require same energy, the first is decomposition and second is combination.
In equation 2, NaOH (s) + H+ + Cl- ------> H2O + Na+ (aq) + Cl- (aq) ?H2
=> Since NaOH(s) heat of formation can be obtained from equation (1), the equation 2 can be re-written as equation3, Na+(aq) + OH-(aq) + H+(aq) + Cl- (aq) ------> H2O + Na+ (aq) + Cl- (aq) ?H3
where the heat of reaction will be sum of the heat of reactions 1 and 2,
?H3 = ?H2 + ?H1 (ignoring the signs )
2) Na+(aq) + OH- (aq) ------> NaOH (s) ?H1 = 44.2 KJ/mol
NaOH (s) + H+ + Cl- ------> H2O + Na+ (aq) + Cl- (aq) ?H2 = -83.602 KJ/mol
Na+(aq) + OH-(aq) + H+(aq) + Cl- (aq) ------> H2O + Na+ (aq) + Cl- (aq)
?H3 = - 44.2 + (-83.602) KJ/mol
= - 127.802 KJ/mol
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