Question

5. A student carries out a calorimetry experiment using HCl and NaOH. What effect will each of the following have on the calculated enthalpy of the reaction relative to the actual enthalpy of reaction? (larger, smaller or no change). Explain your reasoning a. The thermometer always registered a temperature 0.30 °C higher than the actual temperature b. The calorimeter constant used in the calculations by the student is larger than the actual value. (Hint: The best way to answer this question is to take your calorimeter constant, add at least 5 or 10 to the value and quickly redo the calculations. After completing the calculations, you should be able to complete the following statement: As the calorimeter constant increases, the calculated enthalpy of the reaction- --.) c. Unfortunately, the HCI concentration was actually 1.90 M instead of the 2.00 M HCI used in the calculations. (Hint: do a similar calculation. This is an estimate of how errors affect the result. If your data is not matching the expected values, perhaps these questions will help you understand why.) d. The actual concentration of the NaOH prepared for the experiment is 2.00 M instead of the 2.05 M used in the calculations. 6. Calculate the moles of water formed when 25.00 mL of 3.00 M HCl is reacted with 25.00 mL of 3.01 M NaOH. Write a balanced equation.

Answer 1

Question 5:

a) There is no change since the temperature always read more than 0.30C, since the Delta T will still be the same, difference will be (T2+0.30 - T1 - 0.30), which is same as (T2-T1)

b) Delta H(rxn) will be larger since if we use higher specific heat capacity value, then the calculated heat absorbed/released will be more than the expected value, which will lead to higher value of Delta H

c) Delta H will be reduced, since number of moles of HCl will be lesser than the actual value used in the computation

d) Delta H will be increased, since number of moles of NaOH will be higher than the actual value used in the computation

Question 6:

The balanced reaction will be

HCl + NaOH ------------- NaCl + H2O

Number of moles of HCl = Volume of solution (in L) * Molarity (M) = 25.00/1000 * 3.00 = 0.075 moles

Number of moles of NaOH = Volume of solution (in L) * Molarity (M) = 25.00/1000 * 3.01 = 0.07525 moles

Hence the moles of water formed will be governed by the limiting reagent, therefore number of moles of water formed will be 0.075 moles

Note - Post any doubts/queries in comments section for any further explanation.