How would the temperature change and the calculated heats of neutralization vary

Question

How would the temperature change and the calculated heats of neutralization vary if the concentrations of the acids and the base used in this experiment were doubled? Explain

Are the heats of neutralization for the reaction between hydrochloric acid and aqueous sodium hydroxide and for the reaction between nitric acid and aqueous sodium hydroxide the same? Explain why. Would you expect the heats of neutralization for the reactions to be the same if barium hydroxide were used instead of sodium hydroxide? Explain why

Explanation / Answer

Doubling the concentrations of acids and base in the reaction , will basically double up the number of moles participating in the reaction , hence the heat released in the reaction will be doubled and subsequently the temperature change will also be doubled .

Yes the heats of neutralisation in both the reactions will be the same , as the enthalpy change of the neutralisation reaction depends upon the number of moles of water formed in the reaction .

Since both HCl and HNO3 are strong acids , having the same number of protons per mole , they will release the same amount of heat of neutralisation when reacted with strong base i.e. NaOH.

The reaction between Ba(OH)2 and HCl is given as ,

Ba(OH)2 + 2HCl = BaCl2 + 2H2O

So the heat of neutraloisation per mole of water remains same , as there is formation of 2 moles of water in the reaction . But if you see according to number of moles of base, then it is doubled .

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