Question

What will be the heat released when $0.25mole$ of $NaOH$ is titrated against $0.25mole$ $HCl$ for the reaction:
$NaOH + HCl \to NaCl + {H_2}O$
The heat of neutralization for the reaction is $57.1kJ/mol$.
A.$22.5kJ$
B.$57.1kJ$
C.$28.6kJ$
D.$14.3kJ$

Answer 1

We know that, heat of neutralization for the reaction is $57.1kJ/mol$.
To calculate heat of neutralization for $0.25mole$ of $NaOH$ and $HCl$, multiply $0.25mole$ with $57.1kJ/mol$.

Complete step by step answer:
When one equivalent of an acid and one equivalent undergo a neutralization reaction to form water and salt then, there occurs a change in enthalpy. This change in enthalpy is called enthalpy of neutralization. It is denoted by $\Delta {H_n}$. Enthalpy of neutralization can be defined as energy released with the formation of 1 mole of water.
When all strong acids and strong bases are completely ionized in dilute solution then, enthalpy of neutralization is always constant for strong acid and a strong base.
According to the question, it is given that –
The heat of neutralization for the 1 mole of $NaOH$ and $HCl$ is $57.1kJ/mol$.
The reaction given in the question is –
$NaOH + HCl \to NaCl + {H_2}O$
Now, we have to find out the enthalpy of neutralization for $0.25mole$ of $NaOH$ and $HCl$
Therefore, Heat of neutralization –
$\Delta {H_n} = 0.25 \times 57.1 \\\ \Rightarrow \Delta {H_n} = 14.3kJ \\\$
So, the enthalpy of neutralization for $0.25mole$ of $NaOH$ and $HCl$ is $14.3kJ$.
Hence the correct answer is option D.

Note:
When the reaction is carried out under standard conditions at the temperature of 298K and 1 atm of pressure and one mole of water is formed it is called the standard enthalpy of neutralization.
The heat released during the reaction is given by –
$Q = m{S_H}\Delta T$
where, $m$ is the mass of solution
${S_H}$ is the specific heat capacity of solution
$\Delta T$ is the temperature change of the solution.