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Question: When \( 50.0{\rm{ mL}} \) of \( 1.0{\rm{ mol}}{{\rm{L}}^{ - 1}} \) hydrochloric acid is neutralized ...

When 50.0mL50.0{\rm{ mL}} of 1.0molL−11.0{\rm{ mol}}{{\rm{L}}^{ - 1}} hydrochloric acid is neutralized completely by 75.0mL75.0{\rm{ mL}} of 1.0molL−11.0{\rm{ mol}}{{\rm{L}}^{ - 1}} sodium hydroxide in a coffee-cup calorimeter, the temperature of the total solution changes from 20.2oC{20.2^o}C to 25.6oC{25.6^o}C . Determine the quantity of energy transferred, qq and state whether the reaction was endothermic or exothermic.

Explanation

Solution

Hint : Neutralization reaction: When an acid reacts with base to produce water and salt, then this chemical reaction is known as neutralization reaction. The pH of the solution after the reaction, depends on the strength of an acid and a base reacting. If a strong acid is reacted with the strong base, then the solution is said to be completely neutralized and its pH is 77 .

Complete Step By Step Answer:
For the given process, the reaction proceeds as follows:
HCl+NaOH→NaCl+H2OHCl + NaOH \to NaCl + {H_2}O
As per given data,
Concentration of HCl=1.0molL−1HCl = 1.0{\rm{ mol}}{{\rm{L}}^{ - 1}}
Volume of HCl=50mLHCl = 50{\rm{ mL}}
Concentration of NaOH=1.0molL−1NaOH = 1.0{\rm{ mol}}{{\rm{L}}^{ - 1}}
Volume of NaOH=75mLNaOH = 75{\rm{ mL}}
Since the solution of acid and base is very diluted so, we can consider the density and specific heat of the solution to be 1gmL−11{\rm{ gm}}{{\rm{L}}^{ - 1}} and 4.18J(goC)−14.18{\rm{ J(}}{{\rm{g}}^o}C{)^{ - 1}} respectively. Therefore, given volumes of the solutions will approximately equal to their mass.
Assuming the calorimeter to be well insulated, the quantity of heat transferred to the solution will be as follows:
qsolution=mCdT{q_{solution}} = mCdT
Substituting values:
⇒qsolution=(75+50)×4.18×(25.6−20.2)\Rightarrow {q_{solution}} = (75 + 50) \times 4.18 \times (25.6 - 20.2)
⇒qsolution=2821.5J\Rightarrow {q_{solution}} = 2821.5{\rm{ J}}
As we know that, for neutralization process,
qsolution=−qreaction{q_{solution}} = - {q_{reaction}}
Therefore, enthalpy of reaction =−2821.5J= - 2821.5{\rm{ J}}
The negative sign of enthalpy indicates the heat is released during the reaction and hence it is an exothermic reaction.

Note :
A coffee-cup calorimeter is always operated at constant pressure i.e., the heat gained or released in the process is equal to the enthalpy change of the reaction whereas for bomb-calorimeter, the system is at constant volume i.e., the heat gained or released in the process is equal to the internal energy of the system.