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Question: \(150{\text{ ml}}\) of \(0.5{\text{ N}}\) nitric acid solution at \({25.35^ \circ }{\text{C}}\) was ...

150 ml150{\text{ ml}} of 0.5 N0.5{\text{ N}} nitric acid solution at 25.35C{25.35^ \circ }{\text{C}} was mixed with 150 ml150{\text{ ml}} of 0.5 N0.5{\text{ N}} sodium hydroxide solution at the same temperature. The final temperature was recorded to be 28.77C{28.77^ \circ }{\text{C}}. The heat of neutralisation of nitric acid with sodium hydroxide is:
A) 12.64 kcal - 12.64{\text{ kcal}}
B) 11.98 kcal - 11.98{\text{ kcal}}
C) 13.68 kcal - 13.68{\text{ kcal}}
D) 12.68 kcal - 12.68{\text{ kcal}}

Explanation

Solution

The heat of reaction released during the neutralisation of an acid or a base is known as the heat of neutralisation. The neutralisation reaction is an exothermic reaction. in a neutralisation reaction, strong acid and strong base completely dissociates in the solution.

Formulae Used:
Q=mcΔTQ = mc\Delta T
Heat of neutralisation=QMV{\text{Heat of neutralisation}} = \dfrac{Q}{{MV}}

Complete step by step answer:
150 ml150{\text{ ml}} of nitric acid solution at 25.35C{25.35^ \circ }{\text{C}} was mixed with 150 ml150{\text{ ml}} of sodium hydroxide solution at the same temperature.
Thus, the total mass of the solution is,
Total mass of the solution=(150+150) mL{\text{Total mass of the solution}} = \left( {150 + 150} \right){\text{ mL}}
Total mass of the solution=300 mL{\text{Total mass of the solution}} = 300{\text{ mL}}
Thus, the total mass of the solution is 300 mL300{\text{ mL}}.
Calculate the heat required using the equation as follows:
Q=mcΔTQ = mc\Delta T
Where Q is the heat energy,
m is the mass of the solution,
c is the specific heat,
ΔT\Delta T is the change in temperature.
Thus,
Q=300×1×(28.7725.35)Q = 300 \times 1 \times \left( {28.77 - 25.35} \right)
Q=1026 calQ = 1026{\text{ cal}}
Thus, the heat required is 1026 cal1026{\text{ cal}}.
Calculate the heat of neutralisation using the equation as follows:
Heat of neutralisation=QMV{\text{Heat of neutralisation}} = \dfrac{Q}{{MV}}
Where QQ is the heat energy,
N is the normality of the solution,
V is the volume.
Heat of neutralisation=10260.5×150{\text{Heat of neutralisation}} = \dfrac{{1026}}{{0.5 \times 150}}
Heat of neutralisation=0.01368 cal=13.68 kcal{\text{Heat of neutralisation}} = 0.01368{\text{ cal}} = 13.68{\text{ kcal}}
Thus, the heat of neutralisation is 13.68 kcal - 13.68{\text{ kcal}}.

Thus, the correct option is (C) 13.68 kcal - 13.68{\text{ kcal}}.

Note: The negative sign of the value of the heat of neutralisation is because the neutralisation reaction is an exothermic reaction. In an exothermic reaction, heat is released. If the reaction is endothermic and heat is absorbed in the reaction, the value has a positive sign. The complete dissociation of strong acids and strong bases occurs during a neutralisation reaction. A strong covalent bond is then formed between hydrogen and the hydroxide ions to form water molecules. This bond formation process releases a large amount of heat. Thus, the neutralisation reaction is an exothermic process.