When working with the born-haber cycle you can work out questions about working out an enthalpy in the reaction there are different ways you can solve this sort of question. Mathematical equation: ΔH form (NaCl) = ΔH at (Na) + ΔH at (1/2Cl 2 ) + ΔH 1st ion (Na) + ΔH ea (Cl) + ΔH La (NaCl) (or - ΔH Lattice dissociation (NaCl) )
Find the enthalpy change, ΔH1 in kJ/mol (moles of solute = n(NaOH) = mass(NaOH) ÷ M(NaOH) m(NaOH) = 0.5 g M(NaOH) = 22.99 + 16.00 + 1.008 = 39.998 g mol-1 n(NaOH) = 0.5 g ÷ 39.998 g mol-1 =0.013 mol ΔH1 = q ÷ n(NaOH) ΔH1 is negative (process is exothermic) q = -0.523 kJ n(NaOH) = 0.013 mol ΔH1 = -0.523 kJ ÷ 0.013 mol = -40.231 kJ mol-1 Part B : heat of neutralization (NaOH solution + HCl solution) 1.
Confusion over the question is sometimes caused by misconceptions about what heat is. The reason for the lengthy discussions in Lesson 1 was to provide a solid conceptual foundation for understanding the mathematics of Lesson 2.
May 10, 2011 · c) It is exothermic, with a negative enthalpy and positive entropy change. d) It is endothermic, with a positive enthalpy and negative entropy change. 6)4130 J of energy is added to a 52 g sample of water that has an initial temperature of 10.0°C. If the specific heat of water is 4.18 J/(g × °C), what would the final temperature of the water be?