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Question: Equal volume of HCOOH and NaOH are mixed. If x is the heat of formation of water, then heat evolved ...

Equal volume of HCOOH and NaOH are mixed. If x is the heat of formation of water, then heat evolved of neutralization is :

A

More than x

B

Equal to x

C

Twice of x

D

Less than x

Answer

Less than x

Explanation

Solution

The heat of neutralization for a strong acid and a strong base is approximately constant, around -57.3 kJ/mol. This value represents the heat evolved when one mole of H+ ions combines with one mole of OH- ions to form one mole of water:

H+(aq)+OH−(aq)→H2O(l)H^+(aq) + OH^-(aq) \rightarrow H_2O(l)

The question states that 'x' is the heat of formation of water, which implies that 'x' is the magnitude of heat evolved for this reaction (i.e., ΔH=−x\Delta H = -x).

In this problem, we are mixing HCOOH (formic acid) and NaOH (sodium hydroxide). HCOOH is a weak acid, and NaOH is a strong base.

When a weak acid reacts with a strong base, the neutralization reaction occurs in two steps:

  1. Ionization of the weak acid: Since HCOOH is a weak acid, it does not completely dissociate in solution. For the neutralization to proceed, the unionized HCOOH molecules must ionize to provide H+ ions. This ionization process is generally endothermic (requires energy) or at least consumes some energy.

HCOOH(aq)⇌H+(aq)+HCOO−(aq)HCOOH(aq) \rightleftharpoons H^+(aq) + HCOO^-(aq) (ΔHionization>0\Delta H_{ionization} > 0)

  1. Formation of water: The H+ ions produced from the ionization of HCOOH react with the OH- ions from the strong base NaOH to form water. This step is exothermic and releases heat, which is 'x' per mole of water formed.

H+(aq)+OH−(aq)→H2O(l)H^+(aq) + OH^-(aq) \rightarrow H_2O(l) (ΔHwater=−x\Delta H_{water} = -x)

The overall heat of neutralization for the reaction of a weak acid and a strong base is the sum of the enthalpy change of ionization of the weak acid and the enthalpy change of formation of water:

ΔHneutralization=ΔHionization(HCOOH)+ΔHwater\Delta H_{neutralization} = \Delta H_{ionization} (HCOOH) + \Delta H_{water}

Since ΔHionization\Delta H_{ionization} for a weak acid is positive (endothermic), some of the heat released during the formation of water is consumed to ionize the weak acid.

Therefore, the net heat evolved during the neutralization of a weak acid (HCOOH) by a strong base (NaOH) will be less than the heat evolved during the neutralization of a strong acid by a strong base (which is 'x').

In terms of magnitude of heat evolved:

Heat evolved = −ΔHneutralization=−(ΔHionization+ΔHwater)- \Delta H_{neutralization} = - (\Delta H_{ionization} + \Delta H_{water})

Heat evolved = −ΔHionization−ΔHwater- \Delta H_{ionization} - \Delta H_{water}

Since ΔHwater=−x\Delta H_{water} = -x, we have:

Heat evolved = −ΔHionization−(−x)=x−ΔHionization- \Delta H_{ionization} - (-x) = x - \Delta H_{ionization}

As ΔHionization>0\Delta H_{ionization} > 0, the heat evolved will be less than 'x'.