Heat of Neutralization – Weak Acid and Strong Base 1. Complete and balance the f

Question

Heat of Neutralization – Weak Acid and Strong Base

1. Complete and balance the following chemical reaction: NaOH + CH3COOH --> H2O + NaCH3COO ??

2. Volume and Concentration of CH3COOH 50.0 mL 3.0 M

3. Volume and Concentration of NaOH 50.0 mL 3.0 M

4. Calculate the number of moles used of CH3COOH 0.15 mol CH3COOH

0.05 L (3.0 M) =0.15 mol

5. Calculate the number of moles used of NaOH 0.15 mol NaOH

6. Assuming a density of 1 g/mL, what is the total mass of the solution in the calorimeter. ______________ g ??

7. Initial temperature of the CH3COOH solution, Ti 22.2 °C

8. Initial temperature of the NaOH solution, Ti 22.3 °C

9. Final temperature of the solution, Tf 41.7 °C

10. Calculate the change in temperature of the solution, T 19.45 °C

Question:

11. Calculate the heat either released or absorbed by the solution (water)(qrxn) _______________J ??

12. the change in enthalpy (Hrxn) for this reaction _____________J ??

13. Hrxn _____________ kJ ??

14. Which compound is the limiting reagant for this reaction? Calculate the change in enthalpy (Hrxn) per mole of limiting reagant for this reaction. mol (of limiting reagent) ___________mol ??

Hrxn ______________kJ/mol ??

15. If the heat capacity of the calorimeter was 25 J/oC, what would the change in enthalpy (Hrxn) for this reaction be? (again, use equation 7 to recalculate Hrxn.)

__________ kJ/mol ??

16. Is this an exo- or endothermic reaction? Explain the reasoning behind your choice.

17. What is the percent error between your value and the literature value? What are some possible sources of error in the experiment? __________% error ??

Explanation / Answer

NaOH + CH3COOH --> H2O + NaCH3COO

number of moles used of CH3COOH = 0.15 mol

number of moles used of NaOH = 0.15 mol

mass of solution = 100*1 = 100 grams

11. as final temperature is higher than initial temperature..heat is released.

q = m*s*DT   (specific heat of solution = 4.18 j/c.g)

        = 100*4.18*19.45

   heat released = 8130.1 joule

12. DHrxn = 8130.1 joule

13. DHrxn = 8.13 kj

DHrxn = 8.13/0.15 = 54.2 kj/mol

14. both are same no of mol .so that there is no limiting reactant

15. heat absorbed by calorimeter = 25*19.45 = 486.25 j

    DHrxn = 8130.1-486.25 = 7643.85 joule

    DHrxn = 7.643 kj

    DHrxn = 7.643/0.15 = 50.95 kj/mol

16. it is exothermic reaction.

17. literature value not given

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