B. Enthalpy (Heat) of Neutralization for an Acid-Base Reaction HCI + NaOH Trial
ID: 702760 • Letter: B
Question
B. Enthalpy (Heat) of Neutralization for an Acid-Base Reaction HCI + NaOH Trial 2 Trial 1 Trial 1 Trial 2 1. Volume of acid (mL.) 2. Temperature of acid c) 3. Volume of NaOH (mL) 4. Temperature of NaOH C) 5. Exact molar concentration of NaOH (mol/L) 6 Maximum temperature from graph (c) 7. Instructor's approval of graph 2e 2 29.7 025 02 02s 025 305 30 3 29.7 0.l (30.7 ?41aas 307 Calculations for Enthalpy (Heat) of Neutralization for an Acld-Base Reaction d805 as.l asa a-lob 3.3 lade 10 1. Average initial temperature of acid and NaOHCC) 2. Temperature change, AT ec) 3. Volume of final mixture (mL.) 4. Mass of final mixture (g) (Assume the density i051a ??05 | of the solution is 1.0 g/mL) S. Specific heat of mixture 6. Heat evolved 7. Moles of OH reacted, the limiting reactant (mol) 8. Moles of H,O formed (mol) 9. AH, (/mol H,0), equation 25,8 4.18 J/g C 4.18 Jlg C 10. Average AH, (kmol H,0) Show calculations for Trial 1 using the correct number of significant figures. Comment on your two values ofExplanation / Answer
HCl + NaOH = H2O + NaCl
For trial 1
Heat evolved = m x Cp x change in temperature
= 0.05 g x 4.18 J/gC x 2.65 C
= 0.55385 J = 0.00055385 kJ
Moles of OH- reacted = 0.1 mol/L x 0.025 mL x 1L/1000 mL
= 0.0000025 mol
Moles of H2O formed = Moles of OH- reacted
= 0.0000025 mol
Heat of reaction H = 0.00055385 kJ / 0.0000025 mol
= 221.54 kJ/mol
HCl + NaOH = H2O + NaCl
For trial 2
Heat evolved = m x Cp x change in temperature
= 0.05 g x 4.18 J/gC x 3.3 C
= 0.6897 J = 0.0006897 kJ
Moles of OH- reacted = 0.1 mol/L x 0.025 mL x 1L/1000 mL
= 0.0000025 mol
Moles of H2O formed = Moles of OH- reacted
= 0.0000025 mol
Heat of reaction H = 0.0006897 kJ / 0.0000025 mol
= 275.88 kJ/mol
Average H = (221.54 + 275.88)/2 = 248.71 kJ/mol
Same applied for next reaction
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