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Hi, here is a basic summary of what we did in a lab; there were 3 reactions: The procedure: Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. NaOH(s)-> Na+(aq) + OH-(aq) ΔH1=-34.121kJ Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of HCl to form water and an aqueous solution of sodium chloride. NaOH(s) + H+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH2=-83.602kJ Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of HCl to form water an an aqueous solution of sodium chloride. H+(aq) + OH-(aq) + Na+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH3= -50.2kJ The ΔH values were calculated by dividing the heat gained by the number of moles (each reaction had 0.05moles of NaOH) The problem: Net ionic equations for reaction 2 & 3: 2: NaOH(s) + H+(aq) -> H2O + Na+(aq) 3: H+(aq) + OH-(aq) -> H2O i) In reaction 1, ΔH1 represents the heat evolved as solid NaOH dissolves. Look at the net ionic equations for reactions 2 and 3 and make similar statements as to what ΔH2 and ΔH3 represent. ii) Compare ΔH2 with (ΔH1 + ΔH3). Explain in sentences the similarity between these two values by using your answer to #5 above. Attempt at answering: i) Firstly, ΔH2 represents the heat evolved as the hydrogen ion displaces the sodium ion, creating a single displacement reaction. ΔH3 represents the heat evolved as the hydrogen and hydroxide ion form water via a neutralization reaction. ii) ΔH2 is equal to (or supposed to be, this is a source of error while calculating) (ΔH1 + ΔH3). The similarity between these two values is that .. (this is where I get confused!)
Source https://www.physicsforums.com/threads/calorimetry-help-chemistry.399653/
Source https://www.physicsforums.com/threads/calorimetry-help-chemistry.399653/
The conversion of more than one substance reactant into one or more distinct substances, products, and subsequent discussion can be characterized as follows:
Reaction Calculation:
Calculating the Reaction 1:
[tex]NaOH\ (s) \rightarrow Na^+ \ (aq) + OH^- \ (aq) + X_1\ \ KJ ......................... (1)[/tex]
[tex]NaOH[/tex] mass = [tex]1\ g[/tex]
[tex]H_2O[/tex] mass = [tex]50 \ mL = 50\ g[/tex]
water heat of [tex]s_p[/tex] = [tex]4.186\ \frac{ J}{ g\ ^{\circ}C}[/tex]
[tex]\Delta T[/tex] = final temp - initial temp [tex]= 30.3 - 25 = 5.3^{\circ} \ C\\[/tex]
Therefore
Calculating the releasing heat
= mass × sp heat × [tex]\Delta T[/tex]
= 50 × 4.186 × 5.3 J
= 1109.3 J
Calculating the [tex]NaOH[/tex] mass [tex]= 1\ g = \frac{1}{ 40}\ mole= 0.025 \ mole[/tex]
Calculating the releasing heat per mole:
[tex]\to NaOH = \frac{1109.3}{ 0.025} = 44372\ J = 44.4\ KJ[/tex]
Thus
[tex]\to X_1 = 44.4\ KJ[/tex]
Calculating the Reaction 2:
[tex]NaOH \ (s) + H^+\ (aq) + Cl^- \ (aq) \rightarrow Na^+ \ (aq) + Cl^- \ (aq) + H_2O + X_2 \ KJ\\[/tex]
Calculating the net ionic from the equation:
[tex]NaOH\ (s) + H^+\ (aq) \rightarrow Na^+ \ (aq) + H_2O \ (l) + X_2 \ KJ ................................... (2)[/tex]
Calculating the [tex]NaOH[/tex] mass:
[tex]= 1\ g = \frac{1 }{ 40} = 0.025\ mole[/tex]
Calculating the [tex]HCl[/tex] mass:
[tex]= 50\ mL = 50\ g[/tex] [ density = 1 approx]
sp heat of the solution [tex]= 4.186 \frac{J}{g\ ^{\circ}C}[/tex] [ assume the sp heat same as water]
[tex]\Delta T[/tex] = final temp - initial temp [tex]= 36.97 - 25 = 11.97^{\circ} \ C[/tex]
Calculating the releasing heat:
= mass × sp heat × [tex]\Delta T[/tex]
= 50 × 4.186 × 11.97 J
= 2505.3 J
Calculating the releasing heat per mole in [tex]NaOH[/tex]:
[tex]= \frac{ 2505.3 }{ 0.025} = 100212\ J = 100.2 KJ[/tex]
Thus
[tex]X_2 = 100.2 \ KJ[/tex]
Calculating the Reaction 3:
[tex]Na^+ \ (aq) + OH^-\ (aq) + H^+ \ (aq) + Cl^- \ (aq) \rightarrow Na^+\ (aq) + Cl^-\ (aq) + H_2O + X_3\ KJ[/tex]
Calculating the net ionic in the given equation
[tex]H^+ + OH\rightarrow H_2O\ (l) + X_3\ KJ .............................................................. (3)[/tex]
Calculating the volume of [tex]NaOH[/tex]:
[tex]= 25 \ mL\ of\ 1.0\ M = 25 \times \frac{1 }{ 1000} \ mole = 0.025 \ mole[/tex]
Calculating the volume of HCl:
[tex]= 25 \ mL\ of\ 1.0\ M = 25 \times \frac{1 }{ 1000} \ mole = 0.025 \ mole[/tex]
Calculating the total volume
[tex]= 50 \ mL = 50\ g[/tex] { density = 1]
Calculating the sp heat in the solution
[tex]= 4.186 \frac{J}{ g \ ^{\circ} C}[/tex] [ assumed the sp heat is the same as water]
[tex]\Delta T[/tex] = final temp - initial temp [tex]= 31.7- 25 = 6.7^{\circ}\ C[/tex]
Calculating the releasing heat
= mass × sp heat × [tex]\Delta T[/tex]
= 50 × 4.186 × 6.7 J
= 1402.3 J
Calculating the releasing heat per mole in [tex]NaOH[/tex]:
[tex]=\frac{1402.3 }{ 0.025} \ J\\\\= 56092\ J\\\\= 56,09\ KJ[/tex]
Therefore
[tex]X_3 = 56.09 \ KJ\\\\X_1 = 44.4\ KJ\\\\X_2 = 100.2\ KJ\\\\X_3 = 56.09\ KJ\\\\X_2 - [ X_1+ X_3 ] = 100.2 - [44.4 + 56.09]\ = 100.2 - 100.49= -0.29[/tex]
So, the difference is equal to zero.
[tex]\to X_2 = X_1 + X_3[/tex]
This is due to the fact that if we add the reaction (1) and (3) we get the reaction (2)
Calculating the difference percentage:
[tex]= [\frac{0.29 }{100.2} ] \times 100 = 0.29\%[/tex]
The number of joules released in reaction 1 would be 4 times what is released in the calculation if we used 4 g of [tex]NaOH[/tex].
[tex]\to 4 \times 1109.3\ J = 4437.2 \ J\\\\[/tex]
Calculating the [tex]NaOH[/tex] moles [tex]= \frac{4}{40} = 0.1[/tex]
[tex]\to X_1 = \frac{4437.2}{ 0.1} = 44372 \ J = 44.4\ KJ[/tex]
As a result, it has no bearing on the solution's molar heat.
Find out more about the reaction here:
brainly.com/question/17434463