Question

Calculate the enthalpy change when 50 mL of 2 M HCl solution reacts with excess NaOH according to the balanced chemical equation: HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) Given that the specific heat capacity of the solution is 4.18 J/g°C, and the initial and final temperatures of the solution are 25°C and 35°C, respectively. Assume a density of 1 g/mL for the solution.

Answer 1

First, we need to determine the amount of heat absorbed by the solution during the reaction. We can do this using the formula:q = mcTwhere q is the heat absorbed, m is the mass of the solution, c is the specific heat capacity, and T is the change in temperature.Given the density of the solution is 1 g/mL, the mass of the 50 mL solution is:m = 50 mL * 1 g/mL = 50 gThe specific heat capacity  c  is given as 4.18 J/gC, and the change in temperature  T  is:T = T_final - T_initial = 35C - 25C = 10CNow we can calculate the heat absorbed  q :q =  50 g  *  4.18 J/gC  *  10C  = 2090 JNext, we need to determine the moles of HCl in the solution. We can do this using the formula:moles = Molarity * VolumeThe molarity of the HCl solution is given as 2 M, and the volume is 50 mL  0.05 L . Therefore, the moles of HCl are:moles = 2 M * 0.05 L = 0.1 molSince the balanced chemical equation shows a 1:1 ratio between HCl and NaOH, we can assume that 0.1 mol of NaOH also reacts.Now we can calculate the enthalpy change  H  per mole of HCl and NaOH reacted. We can do this using the formula:H = q / molesH = 2090 J / 0.1 mol = 20,900 J/molThe enthalpy change for the reaction between HCl and NaOH is 20,900 J/mol.