Find the enthalpy of neutralization of hcl and naoh. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6 mol dm-3 naoh. the temperature rose from 298 k to 325.8 k. the specific heat capacity is the same as water, 4.18 j/k g.

Approximately .

Assumption: the density of the solution is equal to the density of pure water.

Explanation:

The enthalpy of neutralization is defined as the enthalpy change for each moles of water produced. (Clark, Physical & Theoretical Chemistry, Chemistry Libretexts.)

Each mole of formula units will neutralize one mole of to produce one mole of water. and are available at equal volume and concentration. In other words, there's an equal number of both reactants. All and will react to form water.

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In other words, there are of and each. The two will react to produce of water.

How much heat is released?

Assume that the volume of the liquid is equal to the volume of the solution plus the volume of the solution. That's . Assume that the density of the solution is equal to that of water under room temperature. . The mass of the liquid will be .

Change in temperature:

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Heat that the solution absorbed:

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That will also be the amount of heat released from the reaction if there's no energy loss.

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it is called acid bonding

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