Question

in KJ/Mol the heat evolved when 1 mole of sulphuric acid react with sodium hydroxide is approximately

Answer 1

Answer:

the neutralization of HCl and NaOH, the enthalpy change, often called the enthalpy of reaction is ΔH = -58 kJ/mol

Explanation:

When 50 ml of HCL and NaOH react together,then the heat lost by the reacting HCl and NaOH, therefore q = -2.9 x 103 J.

i.e. it is an exothermic reaction, heat was lost to the water and it got warmer.

50.0 mL of HCl X 1.00 mol HCl = 0.0500 mol HCl

                            1000 mL HCl

The same quantity of base, 0.0500 mole NaOH, was used.

To calculate the energy per mole of acid or base, divide the number of joules by the number of moles.

i.e. molar enthalpy = J/mol = -2.9 x 103 J / 0.0500 mol

                                        = -5.8 x 104 J/mol

                                        = -58000 J/mol

                                        = -58 kJ/mol

Therefore, for the neutralization of HCl and NaOH, the enthalpy change, often called the enthalpy of reaction is ΔH = -58 kJ/mol