Question

2. When a neutralization reaction was carried out using 100.0 ml of 0.7890M NH3 water and 100.0 ml of 0.7940M acetic acid, AT was found to be 4.76 "C The specific heat of the reaction mixture was 4.104 -1K-1 and its density was 1.03 g ml 1. The calorimeter constant was 3.361 K-1 1) Calculate AH neutan for the reaction of NH3 and acetic acid 06) At the end of the experiment, it was discovered that the thermometer had not been calibrated. When it was calibrated, it was found that the thermometer read 0.50 "Clow What etfect would this thermometer reading have on the reported AH neutm calculated in (2) above? ou When the time-temperature data graph was reviewed, it was found that an error had been made in determining AT. Instead of 4.76 "C, ATwas actually 4.70'C. Based on this change only, calculate the correct AH neutzn for the reaction of NH3 and acetic acid 30 v Calculate the percent error for the correct AH neutenifa ATof 4.76 'Chad been used 3. Consider the powbility of substituting a glass beaker for the Styrofoam cup for the calorimeter when doing the experiment described in this module Would you expect the calorimeter constant of the cup to be larger, aller, or the same as that of the beaker Briefly explain ( 2) Would you expect the Arinetan to be larger, smaller, or the same when using the beaker instead of the cup Briefly explain

Answer 1

j* Heat of neutralization — Amount of heat evolved i during neutral, zation of 1 gram equivalent of acid by I gram equivalent of base in a dilute solution. Legé Hnoz (aq) + NaOH(aq) —— Nanoz (aq) + H2000) ; OH = -57.1 KJ Mathematically, - Heat evolved by the neutralization(a)= Mass 04 Solution x Specific heat x (Te-T) Where, T, - Initial tembeostuse of acid and base_ - Ta —— Final temperature of solution after mixing. Enthalpy of heutralization - & looo x1 V 2 Elhere, 2 - Normality of acid or base. ✓ volume of acid or base. Given: 200ml of 0.7890 M NH ₃ Water 100 ml of 007940 M CH3COOH AT: 4076 c , specific heat rabo = 4.104 Tgik Density = 1.03 gmit Total mass of solution Density * Volume 203 * Cloot loo) = loomI NHg Water + loom, CH3COOH = total volume y from co ho Q = 2064x4.104.J_ x 277. 76% A14.760 = 277.76€ Speafic locat - Llallsc 9. = 234 kJ / 2 234 8 240975 - 980.56 call foom equation ® Heat of heutralization - 1234 Xloop x L x 234824.92 - 0.79 = 297246209 KI since, heat is librated , heat of heutralization will be negative. lie Heat of neutralization = -2348-24.97 - 29724679 kJ

" There will be ho effect of theomo metre reading on the rebooted oH heutralization. This is because, the calculatian, Of OH involves OT C change in temperature). (iii) DT = 4.70°c - 27707 K foom So, __ equation 206* 4.104 x 277.7 3 234 774.2 J. _ Heat oj neutoalization : 234774.24 x 1000 min 0.79 2971825.88 kJ. 2 weeks ago Se liv) lo error = ſo Heoror - Shcorrect Ixloo FA 5. DH correct th sHervor : Heat of neutralization corresponding to ste 4.76€ Alcorsest = Heat of heutralization corresponding to sto 4:70°c -) error - 12972467.9 -2971825.87 woo 2971825.8. 2: 0021).