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    Can someone please help me with Enthalpy Change. I really don't get it and it's bugging me, especially with exams around the corner.
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    Do you have any specific examples?
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    (Original post by victoria98)
    Do you have any specific examples?
    Yes,
    It's from a WJEC past paper.
    I'm will screenshot it so apologies if this looks shoddy, :/
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    (Original post by BrynThomas)
    Yes,
    It's from a WJEC past paper.
    I'm will screenshot it so apologies if this looks shoddy, :/
    I apparently can't screenshot it. Anyway, here's what it says.

    "Zac was asked to measure the molar enthalpy change of neutralisation of sodium hydroxide byhydrochloric acid.NaOH(aq) + HCI(aq) NaCl(aq) + H2O(I)He was told to use the following method:• Measure 25.0cm3 of sodium hydroxide solution of concentration 0.970 mol dm−3 into apolystyrene cup.• Measure the temperature of the solution.• Place the hydrochloric acid solution into a suitable container and measure the temperatureof the solution.• When the temperatures of both solutions are equal add 5.00cm3 of hydrochloric acid tothe sodium hydroxide and stir.• Measure the temperature of the mixture.• Keep adding 5.00cm3 portions of hydrochloric acid, until 50.0cm3 have been added,stirring and measuring the temperature each time."
    "((d) Use both values from part (b) to calculate the heat given out during this experiment. [Assume that the density of the solution is 1.00g cm−3 and that its specific heat capacityis 4.18J K−1 g−1] [1]Heat given out = ................................ ........ J(e) Calculate the molar enthalpy change, ∆H, for the reaction between sodium hydroxide andhydrochloric acid. [2]∆H = ................................ ........ kJ mol−1"
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    (Original post by BrynThomas)
    I apparently can't screenshot it. Anyway, here's what it says.

    "Zac was asked to measure the molar enthalpy change of neutralisation of sodium hydroxide byhydrochloric acid.NaOH(aq) + HCI(aq) NaCl(aq) + H2O(I)He was told to use the following method:• Measure 25.0cm3 of sodium hydroxide solution of concentration 0.970 mol dm−3 into apolystyrene cup.• Measure the temperature of the solution.• Place the hydrochloric acid solution into a suitable container and measure the temperatureof the solution.• When the temperatures of both solutions are equal add 5.00cm3 of hydrochloric acid tothe sodium hydroxide and stir.• Measure the temperature of the mixture.• Keep adding 5.00cm3 portions of hydrochloric acid, until 50.0cm3 have been added,stirring and measuring the temperature each time."
    "((d) Use both values from part (b) to calculate the heat given out during this experiment. [Assume that the density of the solution is 1.00g cm−3 and that its specific heat capacityis 4.18J K−1 g−1] [1]Heat given out = ................................ ........ J(e) Calculate the molar enthalpy change, ∆H, for the reaction between sodium hydroxide andhydrochloric acid. [2]∆H = ................................ ........ kJ mol−1"
    Now for b) Temperature Change says it was 6.2 degrees Celcius and Volume of acid required to neutralise Sodium Hydroxide is 26cm^3
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    (Original post by BrynThomas)
    I apparently can't screenshot it. Anyway, here's what it says.

    "Zac was asked to measure the molar enthalpy change of neutralisation of sodium hydroxide byhydrochloric acid.NaOH(aq) + HCI(aq) NaCl(aq) + H2O(I)He was told to use the following method:• Measure 25.0cm3 of sodium hydroxide solution of concentration 0.970 mol dm−3 into apolystyrene cup.• Measure the temperature of the solution.• Place the hydrochloric acid solution into a suitable container and measure the temperatureof the solution.• When the temperatures of both solutions are equal add 5.00cm3 of hydrochloric acid tothe sodium hydroxide and stir.• Measure the temperature of the mixture.• Keep adding 5.00cm3 portions of hydrochloric acid, until 50.0cm3 have been added,stirring and measuring the temperature each time."
    "((d) Use both values from part (b) to calculate the heat given out during this experiment. [Assume that the density of the solution is 1.00g cm−3 and that its specific heat capacityis 4.18J K−1 g−1] [1]Heat given out = ................................ ........ J(e) Calculate the molar enthalpy change, ∆H, for the reaction between sodium hydroxide andhydrochloric acid. [2]∆H = ................................ ........ kJ mol−1"
    Ok so to work out the heat given out, you need to use the q=mcΔT formula where m is the mass of solution (26 + 25 in this case). q=51 x 4.18 x 6.2 = 1321.7 J

    Enthalpy change is given by q/n. You can work out the number of moles of NaOH - comes out at 0.0243 moles Heat was given OUT (temperature ROSE) so the reaction is exothermic. ΔH = (1321.7 x 10^-3)/0.0243=-54.5kJmol^-1

    Which part do you not get?
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    Is it actually just that? I've been looking at the mark scheme for ages and that just did not click with me. Bloody hell, I feel like an idiot now. But yeah, that makes sense. Cheers for that.

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