Question

If the heat of neutralization for a strong acid-base reaction is $- 57.1{\text{kJ}}$ , what would be the heat released when $350{\text{c}}{{\text{m}}^{\text{3}}}$ of $0.20$ M ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is mixed with 650 ${\text{c}}{{\text{m}}^{\text{3}}}$ of $0.10$ M ${\text{NaOH}}$ ?
A.${\text{8}}{\text{.5kJ}}$
B.${\text{17}}{\text{.1kJ}}$
C.${\text{5}}{\text{.7kJ}}$
D.${\text{3}}{\text{.71kJ}}$
E.${\text{2}}{\text{.85kJ}}$

Answer 1

The enthalpy of neutralization is defined as the change in enthalpy when 1 mole of hydrogen ion in dilute solution combines with 1 mole of hydroxide ion to give rise to undissociated water. The heat released for ‘n’ number of moles can thus be determined from the enthalpy of neutralization by taking the product of ‘n’ with the enthalpy of neutralization.

Complete step by step answer:
According to the given question, the heat of neutralization, i.e., the enthalpy of neutralization is found to be $- 57.1{\text{kJ}}$ . This enthalpy of neutralization of $- 57.1{\text{kJ}}$ corresponds to the reaction shown below.
${{\text{H}}^{\text{ + }}}\left( {{\text{aq}}} \right){\text{ + O}}{{\text{H}}^{\text{ - }}}\left( {{\text{aq}}} \right) \to {{\text{H}}_2}{\text{O}}\left( {\text{l}} \right)$
This means that the change in enthalpy when one mole of sulphuric acid (which is a strong diprotic acid) is mixed with one mole of sodium hydroxide (which is a strong monoacidic base) is of the order of $- 57.1{\text{kJ}}$ .
We need to find out the amount of heat released when $350{\text{c}}{{\text{m}}^{\text{3}}}$ of $0.20$ M ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is mixed with 650 ${\text{c}}{{\text{m}}^{\text{3}}}$ of $0.10$ M ${\text{NaOH}}$ . So, at first, we need to determine the number of moles of sulphuric acid and sodium hydroxide which are involved in this case.
Let us first calculate the number of moles of sulphuric acid. The amount or moles of a substance is given by the product of its molarity with volume. So, the amount of sulphuric acid, ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is:
$= 0.20 \times 350 \\\ = 70{\text{mmol}} \\\$
But it is a diprotic acid and so there are 2 hydrogen ions. So, the amount of hydrogen ions is:
$= 2 \times 70{\text{mmol}} \\\ {\text{ = 140mmol}} \\\$
Similarly, the amount of sodium hydroxide is:
$= 0.10 \times 650 \\\ = 65{\text{mmol}} \\\$
Hence, the amount of hydroxide ions is $= 65{\text{mmol}}$ .
Thus, we can see that sodium hydroxide is the limiting reactant and therefore, 65 mmol of hydroxide ions will combine with 65 mmol of hydrogen ions to produce 65 mmol of water.
1 mol of hydrogen ions reacts with 1 mole of hydroxide ions and the production of $57.1{\text{kJ}}$ occurs in this process.
Thus, the heat produced will be the product of the enthalpy of neutralization with 65 mmol or $65 \times {10^{ - 3}}$ moles.
Hence, the heat released when $350{\text{c}}{{\text{m}}^{\text{3}}}$ of $0.20$ M ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is mixed with 650 ${\text{c}}{{\text{m}}^{\text{3}}}$ of $0.10$ M ${\text{NaOH}}$ is
$= 57.1 \times 65 \times {10^{ - 3}}{\text{kJ}} \\\ {\text{ = 3}}{\text{.71kJ}} \\\$

So, the correct option is D.

Note: When a strong acid and a weak base, or a weak acid and a strong base or a weak acid and a weak base are mixed in equivalent amounts, then the heat change is less than $- 57.1{\text{kJ}}$ or $- 13.7{\text{kcalmo}}{{\text{l}}^{ - 1}}$. For example, for the following reaction the heat change is $- 12.3{\text{kcalmo}}{{\text{l}}^{ - 1}}$ .
${\text{HCl}}\left( {{\text{aq}}} \right) + {\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}\left( {{\text{aq}}} \right) \to {\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\left( {{\text{aq}}} \right) + {{\text{H}}_2}{\text{O}}\left( {\text{l}} \right)$
This happens because a portion of the heat evolved is utilized in the complete ionization of a weak acid or a weak base or both. So, the net heat of neutralization is less.