Question

Calculate the standard enthalpy change (∆H°) for the dissolution of 5.00 g of solid NaOH in 100.0 mL of water, given that the temperature of the solution increased from 25.0°C to 27.5°C. The density of the solution is 1.00 g/mL and the specific heat capacity of the solution is 4.18 J/(g°C). The molar mass of NaOH is 40.00 g/mol.

Answer 1

To calculate the standard enthalpy change  H  for the dissolution of NaOH, we will use the formula:q = mcTwhere q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity, and T is the change in temperature.First, we need to find the mass of the solution. Since the density of the solution is 1.00 g/mL, the mass of 100.0 mL of water is:mass of water = density  volume = 1.00 g/mL  100.0 mL = 100.0 gNow, we add the mass of NaOH to the mass of water to find the total mass of the solution:mass of solution = mass of water + mass of NaOH = 100.0 g + 5.00 g = 105.0 gNext, we find the change in temperature:T = final temperature - initial temperature = 27.5C - 25.0C = 2.5CNow, we can calculate the heat absorbed or released  q  using the formula:q = mcT =  105.0 g  4.18 J/ gC   2.5C  = 1096.95 JSince the temperature of the solution increased, the dissolution of NaOH is exothermic, and the heat released  q  will be negative:q = -1096.95 JNow, we need to find the moles of NaOH dissolved:moles of NaOH = mass of NaOH / molar mass of NaOH = 5.00 g / 40.00 g/mol = 0.125 molFinally, we can calculate the standard enthalpy change  H  for the dissolution of NaOH:H = q / moles of NaOH = -1096.95 J / 0.125 mol = -8775.6 J/molThe standard enthalpy change for the dissolution of 5.00 g of solid NaOH in 100.0 mL of water is -8775.6 J/mol.