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 C low. What effect would thus thermometer reading have on the reported change in H neoutzn calculated above? **Text from the Image:**© 1988 Cengage Learning2. When a neutralization reaction was carried out using 100.0 mL of 0.7980 M NH₃ water and 100.0 mL of 0.7940 M acetic acid, ΔT was found to be 4.76°C. The specific heat of the reaction mixture was 4.104 J/g·K, and its density was 1.03 g/mL. The calorimeter constant was 3.36 J/°C.(1) Calculate ΔH_neutzn for the reaction of NH₃ and acetic acid. - heat absorbed by reaction = (volume)(density)(ΔT)(specific heat) = (200 mL)(1.03 g/mL)(4.76)(4.104 J/g·K) = 400.424 J- heat absorbed by calorimeter = calorimeter constant(ΔT) = 3.36 J/°C(4.76 °C) = 15.99 J- total heat = 400.424 J + 15.99 J = 416.423 J- moles of NH₃ = (0.100 L)(0.7980 M) = 0.0798 mol- moles of CH₃COOH = (0.100 L)(0.7940 M) = 0.0794 mollimiting reagent → ignore ΔH_neutrn = total heat / molesΔH_neutrn = 416.423 J / 0.0794 mol = Fill kJ/mol(2) 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 0.5°C low. What effect would this thermometer reading have on the reported ΔH_neutzn calculated in (2) above?---Note: The calculation for ΔH_neutrn is not completed in the image and is marked as "Fill kJ/mol."

Enthalpy of reaction, in calorimetry, is calculated by the formula: ∆H = mass * specific heat capacity * ∆T Here, ∆T is the temperature change.Answer 2 The thermometer read 0.50 oC low. It means it had an error and actual temperature will be 0.50 oC higher. In present case, the initial and final temperatures are measured which will be 0.50 oC lower. Let the initial temperature measured = T1 oC Actual initial temperature measured = (T1+0.50) oC The final temperature measured = T2 oC Actual final temperature measured = (T2+0.50) oC Change = (T2+0.50) oC - (T1+0.50) oC = (T2 - T1) oC The change remains inspite of the error. The calculation of enthalpy requires change and not actual values, so, there will be no change in calculated ∆H.