Question

Calculate the enthalpy change for the precipitation reaction between 25.0 mL of 0.100 M silver nitrate (AgNO3) and excess calcium chloride (CaCl2) solution at 25°C. The balanced equation for the reaction is:AgNO3(aq) + CaCl2(aq) → AgCl(s) + Ca(NO3)2(aq)Assume the density of the solutions is equal to 1.00 g/mL and the solutions have the same specific heat as water (4.18 J/g°C).

Answer 1

To calculate the enthalpy change for the precipitation reaction, we first need to determine the moles of AgNO3 reacting and then use the enthalpy of formation values for the reactants and products.1. Calculate the moles of AgNO3:moles = Molarity  Volumemoles of AgNO3 = 0.100 mol/L  0.025 L = 0.00250 mol2. Use the balanced equation to determine the stoichiometry of the reaction:1 mol AgNO3 reacts with 1 mol CaCl2 to produce 1 mol AgCl and 1 mol Ca NO3 2.3. Look up the standard enthalpy of formation  Hf  values for the reactants and products:Hf  AgNO3  = -124.4 kJ/molHf  CaCl2  = -795.4 kJ/molHf  AgCl  = -127.0 kJ/molHf  Ca NO3 2  = -814.0 kJ/mol4. Calculate the enthalpy change for the reaction  Hrxn  using the following equation:Hrxn =  [Hf  products ] -  [Hf  reactants ]Hrxn = [ 1 mol AgCl  -127.0 kJ/mol  +  1 mol Ca NO3 2  -814.0 kJ/mol ] - [ 1 mol AgNO3  -124.4 kJ/mol  +  1 mol CaCl2  -795.4 kJ/mol ]Hrxn =  -127.0 kJ + -814.0 kJ  -  -124.4 kJ + -795.4 kJ Hrxn =  -941.0 kJ  -  -919.8 kJ Hrxn = -21.2 kJ/mol5. Calculate the enthalpy change for the given amount of AgNO3:H = Hrxn  moles of AgNO3H = -21.2 kJ/mol  0.00250 molH = -0.0530 kJThe enthalpy change for the precipitation reaction between 25.0 mL of 0.100 M silver nitrate and excess calcium chloride solution at 25C is -0.0530 kJ.