Question

reaction one NaOH and HCl. 50.0 mL of 2.0 M HCl solution into 50.0 mL of 2.0 M NaOH

reaction two 50 mL OF 2.0 M NaOH into 50 mL of 2.0 M NH4Cl

reaction three 50.0 mL of 2.0 M HCl into 50.0 mL NH3

pleanse answer all of the questions

Experinen DATA TABLE Reaction 1 Reaction 2 Reaction 3 34,012。21 2H0 Maximum temperature (C) Initial temperature (C) Temperature change (&T) DATA ANALYSIS 1. Calculate the amount of heat energy, g. produced in each reaction. Use 1.03 g/mL. for the density of all solutions. Use the specific heat of water, 4.18 J/(g.°C), for all solutions. 2. Calculate the enthalpy change, AH, for each reaction in terms of kJ/mol of each reactant. 3. Use your answers from 2 above and Hess's law to determine the experimental molar enthalpy for Reaction 3 4. Use Hess's law, and the accepted values of ΔH in the Pre-Lab Exercise to calculate the ΔH for Reaction 3. How does the accepted value compare to your experimental value? 5. Does this experimental process support Hess's law? Suggest ways of improving your results. 13-4 Advanced Chemistry with Vernier

Answer 1

Volume of solution, V = (50 + 50) = 100 ml mass, m = (100 ml) (1.03 g mol^-1) = 103 g Thus heat energy produced, theta_1 = m C_p dT = (103 g) (4.18 H. g^-1. degree C^-1) (13.3 degree C) = 5726.182 J total volume, V = (50 + 50) = 100 ml mass, m = (100 ml) (1.03 g middot ml^-1) = 103 g heat produced, Q_2 = m C_P d = (103) (4.18) (1.2) = 947 . 188 J Reaction 3 total volume, V = (50 + 50) = 100 ml ram, m = (100) (1.03 g. ml^-1) = 103 g heat produced, Q_3 = (103) (4.18) (17.5) = 7534.45 J \r\n Reaction -1: moles of HCl, n = 50/1000 L middot 2 mol middot L^-1 = 0.1 mol Enthalpy change H = Q_1/n = 5726.182/0.1 J mol^-1 = 57261.82 J middot mol^-1 = 57.26182 KJ middot mol^-1 Reaction 2 moles of N as H