Question

Heat of neutralization for the reaction, is $57.1kJmo{{l}^{-1}}$. The heat released when $0.25$ moles of $NaOH$ is titrated against $0.25$mole $HCl$ will be:
$NaOH+HCl\to NaCl+{{H}_{2}}O$
A.) $22.5kJ$
B.) $57.1kJ$
C.) $28.6kJ$
D.) $14.3kJ$

Answer 1

Hint: First of all balance the chemical reaction and find out the heat of neutralization for the balanced reaction. Now adjust the stoichiometry in such a way that it should satisfy the given moles of the reactants and also the chemical reaction should be balanced, then find out the heat of neutralization for that reaction by manipulating the original value in the same way as how the reaction is manipulated.

Complete Solution :
The given standard equation of the neutralization reaction between the chemical compounds $NaOH$ and $HCl$ is $NaOH+HCl\to NaCl+{{H}_{2}}O$
And the Heat of neutralization for the reaction, is $57.1kJmo{{l}^{-1}}$

- The above equation of the reaction is balanced equation so that we can understand that for one mole of $NaOH$ titrated against one mole of $HCl$ gives $57.1kJmo{{l}^{-1}}$ as the Heat of neutralization for the reaction.
But given to find Heat of neutralization for the reaction for $0.25$ moles of $NaOH$ titrated against $0.25$ moles of $HCl$ so multiply the balanced equation with $0.25$ which gives
$0.25NaOH+0.25HCl\to 0.25NaCl+0.25{{H}_{2}}O$

- We are having Heat of neutralization for the reaction, is $57.1kJmo{{l}^{-1}}$ for one mole each compound is titrated against each other in the reaction, therefore for $0.25$ moles of each compound is titrated against each other in the reaction the Heat of neutralization for the reaction will be obtained by multiplying the given Heat of neutralization for the reaction with $0.25$
That is
$0.25NaOH+0.25HCl\to 0.25NaCl+0.25{{H}_{2}}O;\Delta H=57.1\times 0.25kJmo{{l}^{-}}$
$0.25NaOH+0.25HCl\to 0.25NaCl+0.25{{H}_{2}}O;\Delta H=14.275kJmo{{l}^{-}}$

Therefore for The heat released when $0.25$ moles of $NaOH$ is titrated against $0.25$ mole $HCl$ is obtained as $14.275 kJmo{{l}^{-}}$
So, the correct answer is “Option D”.

Note: When we are finding the heat of formation or the heat of neutralization we should be careful of the stoichiometry and the equation should be balanced so that no errors will take place. Also we should take care that the reactants in the reaction should be in their standard form.