Question

CHMISILLE THERMOCHEMISTRY 19 Name Data / vial + NaOH mass / Partner Reaction 1 Reaction 2 Reaction 3 NaOH(s) + H2O( NaOH(aq) + HCl(aq) NaOH(s) + HCl(aq) 8.6068 8 NA 18.6067 B 6.6314 B NA 16.5960 100.1 mL H0 51.Z ml 1.0 M HCI 999mL. 0.5 M HCI / empty vial mass measured volume / liquid mass (V x density) NaOH used soln mass = liquid + solid 1ov.lg 52.2249 101.8989 1.9727 850./ML 1.0 M NaOH 2.0107 102.0727 8 103.326 8 103.9087 8 30.8 C 31.1 °C 34.5 °C 23.7 °C 24.3 C 23.5 °C 7.1 °C 6.8 maximum temp, 72 initial temp. T) °c temp difference, AT Show calculation details on next page k 4.7834 kJ 3.0329 kJ 2.9405 heat absorbed q = m-sAT assion sign) enthalpy change, AH (assign sign) - kl kJ/mol NaOH kJ/mol NaOH kJ/mol NaOH molar enthalpy of rxn, AH. % error = |(AH+2-AHYAH | 100% PAGE 9 OF 16 CHM151 THERMOCHEMISTRY REV 2019-05-15

Is the enthalpy change just the opposite of the heat absorbed? Do you use that opposite number to calculate the molar enthalpy?

Answer 1

Yes the enthalpy change, $\large \Delta$H is opposite of of the heat absorbed, q. So,

$\large \Delta$H = -q

The molar enthalpy is calculated by dividing $\large \Delta$H by the number of moles of the substance. Thus,

Molar enthalpy = $\large \Delta$H/moles of substance

= -q/moles of substance

Corrections: Reaction 2: A Han = - moles of Naoh -2.9404515 kJ 0.050 I moles = 1-58.6916 kJ/mol Reaction -3! Amen --q moles of NaoH - - 24-783437 kJ 0.0493175 moles =1-96-99269 kJ/mol