Question

Calculate the standard enthalpy change ($\Delta H^\circ$) for the dissolution of 10.0 g of sodium hydroxide (NaOH) in 100.0 mL of water assuming that the specific heat capacity of the solution is 4.18 J g$^{-1}$ K$^{-1}$ and that the initial and final temperatures are 25.0$^\circ$C and 35.0$^\circ$C, respectively. The molar mass of NaOH is 40.0 g mol$^{-1}$.

Answer 1

To calculate the standard enthalpy change for the dissolution of NaOH, we first need to determine the heat absorbed by the solution  q  using the formula:q = mcTwhere 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 mass of NaOH is 10.0 g and the volume of water is 100.0 mL, we can assume that the mass of water is approximately 100.0 g  since the density of water is approximately 1 g/mL . Therefore, the mass of the solution is:m_solution = m_NaOH + m_water = 10.0 g + 100.0 g = 110.0 gNext, we can calculate the change in temperature:T = T_final - T_initial = 35.0C - 25.0C = 10.0CNow we can calculate the heat absorbed by the solution:q = mcT =  110.0 g  4.18 J g$^{-1}$ K$^{-1}$  10.0C  = 4596 JNow we need to find the moles of NaOH:moles_NaOH = mass_NaOH / molar_mass_NaOH = 10.0 g / 40.0 g mol$^{-1}$ = 0.25 molFinally, we can calculate the standard enthalpy change for the dissolution of NaOH:H = q / moles_NaOH = 4596 J / 0.25 mol = 18384 J/molThe standard enthalpy change for the dissolution of NaOH is 18,384 J/mol.