Question

Calculate the enthalpy change of precipitation when 50 ml of 0.1 M silver nitrate (AgNO3) is added to 50 ml of 0.1 M sodium chloride (NaCl) solution. The temperature of the mixture increases by 5.0 °C. Given the specific heat capacity of the solution is 4.18 J/g °C and the density of the solution is 1.0 g/mL.

Answer 1

To calculate the enthalpy change of precipitation, we first need to determine the amount of heat released during the reaction. We can do this using the formula:q = mcTwhere q is the heat released, m is the mass of the solution, c is the specific heat capacity, and T is the change in temperature.First, let's find the mass of the solution. Since the density of the solution is 1.0 g/mL, and we have 50 mL of each solution, the total mass of the solution is:m =  50 mL + 50 mL  * 1.0 g/mL = 100 gNow, we can plug the values into the formula:q =  100 g  *  4.18 J/g C  *  5.0 C  = 2090 JNext, we need to determine the moles of the precipitate formed. The balanced chemical equation for the reaction is:AgNO3  aq  + NaCl  aq   AgCl  s  + NaNO3  aq Since both solutions have the same concentration  0.1 M  and volume  50 mL , they will react in a 1:1 ratio. To find the moles of the precipitate  AgCl  formed, we can use the formula:moles = Molarity * Volumemoles of AgCl = 0.1 M * 0.050 L = 0.005 molFinally, we can calculate the enthalpy change of precipitation  H  using the formula:H = q / molesH = 2090 J / 0.005 mol = 418000 J/molThe enthalpy change of precipitation for this reaction is 418 kJ/mol.