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CHEM201 Molar Enthalpy of Neutralization Pre-Lab Questions student determined AHasuen for the reaction of sodium hydroxide and acetic acid, using the procedure described in this module. The student added 100.0 mL of 0.8500 M NaOH to 100.0 ml. of 0.8404 M acetic acid at 5 minutes. Prior to and following the mixing of the acid and base solutions, the following temperature-time (reference the graphing guide as needed) to answer the following questions. Show wo data were collected. Plot the data rk (with units) to all questions! Time min NaOH emperature, oC Time min mixture 6 23.28 7 23.03 8 23.10 9 23.06 10 22.99 11 22.93 12 22.89 13 22.85 14 22.78 15 22.73 16 22.70 17 22.64 18 22.59 Temperature, °C Temperature, °C 1 17.52 0.5 1.0 17.54 HOAc 17.73 17.75 2.0 17.55 2.5 3.0 17.57 17.78 4.0 17.58 4.5 5.0 mixing 17.80 mixing Plot the data and add trendlines. Attach a copy of the graph to your pre-lab. From the graph, determine ΔT a. b. i. Determine the temperatures of the unmixed reactants right before mixing (x 5 min) using the trendlines of the graph. T456412 Using the values in part i·calculate average temperature of the unmixed reactants (right before mixing). ii. ii. Use the answer to part ii as your initial temperature to calculate AT of the reaction mixture. Note: you need to calculate the final temperature.

Answer 1

a) The plot of temperature (°C) vs time for the mixture of NaOH and HOAc are shown below.

bi) E1 gives the variation of temperature vs time for NaOH (upto 5 min); E2 gives the variation of temperature vs time for HOAc (upto 5 min) and E3 gives the variation of temperature vs time for the mixture (from 6-18 minutes).

Find the temperatures of NaOH and HOAc at 5 minutes.

NaOH: Put x = 5 in E1 and obtain

y = 0.021*5 + 17.5 = 17.605

HOAc: Put x = 5 in E2 and obtain

y = 0.017*5 + 17.724 = 17.809

The temperatures of NaOH and HOAc before mixing are 17.605°C and 17.809°C (ans).

ii) Average temperature of the reactants before mixing = ½*(17.605 + 17.809)°C = 17.707°C (ans).

iii) The final temperature of the mixture can be obtained by extrapolating E3 to zero. Put x = 0 and obtain

y = -0.0504x + 23.495 = -0.504*0 + 23.495 = 23.495

The final temperature of the mixture is 23.495°C.

ΔT = T_final – T_initial = 23.495°C - 17.707°C = 5.788°C (ans).