Part 2: Measuring the Enthalpy of Neutralization 148 Part II: Measuring the Enth

Question

Part 2: Measuring the Enthalpy of Neutralization

148 Part II: Measuring the Enthalpy of Neutralization Trial 1 Trial 2 Molarity NaOH: Volume NaOH: 5Oml Molarity HCI(aq)' d Volume HCI(aq): ? o m I Mole of HC) neutralized: 0.1 Ti Molarity NaOH:2M Volume NaOH: son Molarity HCl 940 Volume HCl Mole of HCleag) neutralized: O.l AT2 Calculation of Observed AHneu Calculation of Observed ??.eut. Observed ??,out- Observed AHeut Calculation of AHneut per mole: Calculation of AHneut per mole: ??neut per mol AHneut per mole = Average of Trials 1 and 2: AHneut per mole

Explanation / Answer

TRIAL 1.

Enthalpy of neutralisation ?H = - Q(energy released during reaction)

Q = m× cp ×?T

cp is the specific heat of solution = SH = 4.184 J/(g.?)

m is mass of solution, since the specific heat of solution is equal to specific heat of water.

So mass of solution should be equal to mass of water.

Density of water is 1 g/ml

This means 1 ml water has mass = 1g

Now, total volume of solution = 50+50=100 ml

So, 100 ml solution has mass = 100 g

Now Q = 100 g × 4.184 J/(g.?) × 12? =50208 J

So, ?H = - 50,208 J

Per mol

?H (neutralisation) = -Q/n

n is number of moles = 0.1 mol

?H per mol = - 50,208J/0.1 mole

= - 502080 J/mol

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