
The heat of neutralization is higher for
(A) $N{{H}_{4}}OH+C{{H}_{3}}COOH$
(B) \[HN{{O}_{3}}+N{{H}_{4}}OH\]
(C) \[NaOH+C{{H}_{3}}COOH\]
(D) \[HCl+NaOH\]
Answer
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Hint: The heat of neutralization would be higher when complete ionization takes place. This will happen on the complete dissociation of strong acid and base.
Complete step by step solution:
The heat of neutralization is also known as enthalpy of neutralization. As we know it is the energy released when one equivalent of acid reacts with one equivalent of the base (neutralization reaction) to produce salt and water as products.
- A strong acid is one which gets dissociated 100 % into ions. A weak acid is one which doesn’t ionize fully. We can similarly explain the strength of bases also in this manner. The strength of acids and bases can also be explained in terms of dissociation constant and pH.
- The Heat of neutralization of strong acid and strong base is always greater than the heat of neutralization of weak acid or weak base. The reason is that some energy is used to ionize weak acids or weak bases.
- In the question, we are given three acids and two bases. So, let's classify them into strong acids, weak acids, strong base and weak base.
$HCl$ - Hydrochloric acid - Strong acid
$C{{H}_{3}}COOH$-Acetic acid -Weak acid
$HN{{O}_{3}}$- Nitric acid- Strong acid
$NaoH$-Sodium Hydroxide- Strong base
$N{{H}_{4}}OH$-Ammonium Hydroxide-Weak base
-Thus, we can see that $HCl$and $NaoH$are the strong acids and strong bases respectively. So, their heat of neutralization will be the highest one among the given options.
-The equation for the reaction between $HCl$and $NaoH$can be given as
$NaO{{H}_{(aq)}}+HC{{l}_{(aq)}}\to NaC{{l}_{(aq)}}+{{H}_{2}}{{O}_{(l)}}$
Therefore, the heat of neutralization is higher for option (D) \[HCl+NaOH\].
Additional Information:
The enthalpy change of neutralization is always measured per moles of water formed.
Note: We need to note down that enthalpy change of neutralization is always negative. It's because heat is given out when alkali and acid react. When strong acids and alkalis are involved in reactions, the values are always very closely similar with values between -57 and -58 k J$mo{{l}^{-1}}$.
Complete step by step solution:
The heat of neutralization is also known as enthalpy of neutralization. As we know it is the energy released when one equivalent of acid reacts with one equivalent of the base (neutralization reaction) to produce salt and water as products.
- A strong acid is one which gets dissociated 100 % into ions. A weak acid is one which doesn’t ionize fully. We can similarly explain the strength of bases also in this manner. The strength of acids and bases can also be explained in terms of dissociation constant and pH.
- The Heat of neutralization of strong acid and strong base is always greater than the heat of neutralization of weak acid or weak base. The reason is that some energy is used to ionize weak acids or weak bases.
- In the question, we are given three acids and two bases. So, let's classify them into strong acids, weak acids, strong base and weak base.
$HCl$ - Hydrochloric acid - Strong acid
$C{{H}_{3}}COOH$-Acetic acid -Weak acid
$HN{{O}_{3}}$- Nitric acid- Strong acid
$NaoH$-Sodium Hydroxide- Strong base
$N{{H}_{4}}OH$-Ammonium Hydroxide-Weak base
-Thus, we can see that $HCl$and $NaoH$are the strong acids and strong bases respectively. So, their heat of neutralization will be the highest one among the given options.
-The equation for the reaction between $HCl$and $NaoH$can be given as
$NaO{{H}_{(aq)}}+HC{{l}_{(aq)}}\to NaC{{l}_{(aq)}}+{{H}_{2}}{{O}_{(l)}}$
Therefore, the heat of neutralization is higher for option (D) \[HCl+NaOH\].
Additional Information:
The enthalpy change of neutralization is always measured per moles of water formed.
Note: We need to note down that enthalpy change of neutralization is always negative. It's because heat is given out when alkali and acid react. When strong acids and alkalis are involved in reactions, the values are always very closely similar with values between -57 and -58 k J$mo{{l}^{-1}}$.
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