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What is the heat of neutralization for the reaction between 50 mL of 1.0 M hydrochloric acid (HCl) and 50 mL of 1.0 M sodium hydroxide (NaOH) at a room temperature of 25°C? The enthalpy of neutralization of HCl and NaOH is -57.1 kJ/mol.

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To calculate the heat of neutralization for the reaction between 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH, we first need to determine the number of moles of each reactant.Moles of HCl =  volume in L    concentration in mol/L Moles of HCl =  0.050 L    1.0 mol/L  = 0.050 molMoles of NaOH =  volume in L    concentration in mol/L Moles of NaOH =  0.050 L    1.0 mol/L  = 0.050 molSince the moles of HCl and NaOH are equal, the reaction goes to completion, and all of the acid and base are neutralized.The enthalpy of neutralization for HCl and NaOH is -57.1 kJ/mol. This means that 57.1 kJ of heat is released for every mole of HCl and NaOH that react. To find the heat of neutralization for this specific reaction, we multiply the enthalpy of neutralization by the moles of HCl  or NaOH, since they are equal .Heat of neutralization =  moles of HCl    enthalpy of neutralization Heat of neutralization =  0.050 mol    -57.1 kJ/mol  = -2.855 kJThe heat of neutralization for the reaction between 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH at 25C is -2.855 kJ.

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