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Calculate the standard enthalpy change for the reaction between 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH at 25°C given that the temperature of the resultant solution is 31.2°C and the specific heat capacity of the solution is 4.184 J/g°C.

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ago by (350 points)
To calculate the standard enthalpy change for the reaction, we first need to determine the heat released or absorbed during the reaction. We can do this using the formula:q = mcTwhere q is the heat released or absorbed, 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 water is approximately 1 g/mL, the mass of the 100 mL solution  50 mL HCl + 50 mL NaOH  is approximately 100 g.Next, we can calculate the change in temperature  T  as follows:T = T_final - T_initial = 31.2C - 25C = 6.2CNow we can calculate the heat released or absorbed  q  using the formula:q = mcT =  100 g  4.184 J/gC  6.2C  = 2594.08 JSince the temperature of the solution increased, the reaction is exothermic, and the heat released  q  is negative:q = -2594.08 JNow we need to determine the moles of the limiting reactant. Both HCl and NaOH have a concentration of 1.0 M and a volume of 50 mL, so they react in a 1:1 ratio. Therefore, the limiting reactant is either one, and we can calculate the moles as follows:moles =  1.0 mol/L  0.050 L  = 0.050 molFinally, we can calculate the standard enthalpy change  H  by dividing the heat released  q  by the moles of the limiting reactant:H = q / moles = -2594.08 J / 0.050 mol = -51881.6 J/molThe standard enthalpy change for the reaction between 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH at 25C is approximately -51.9 kJ/mol.

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