In a calorimetry experiment to measure the heat of neutralization (in joules per mole)...25.0 mL of 2.0 HCL reacts with 25.0 mL of 2.0 M NaOH. Vtotal= 50 mL. The temperature of the solution goes...

In a calorimetry experiment to measure the heat of neutralization (in joules per mole)...25.0 mL of 2.0 HCL reacts with 25.0 mL of 2.0 M NaOH. Vtotal= 50 mL. The temperature of the solution goes from 18 degrees Celsius to 30 degrees celsius. The specific heat of H2O is 4.18 joules/ degree gram. Calculate delta H in joules per mole

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llltkl | College Teacher | (Level 3) Valedictorian

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Assuming that the resulting dilute salt solution have a density of 1gm/ml, total heat evolved is H= ms[T2-T1]

=50x4.18[30-18]Joules =2508 J = 2.508 KJ

Now this amount of heat has been generated in a reaction involving 25x2.0 = 50 meq of acid/base.

So 1000 meq = 1eq acid/base should involveĀ 2.508 x 1000/50 KJ = 50.16 KJ. This is 'Delta' H per mole for this neutralisation reaction.

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