
When 0.2 N \[{H_2}S{O_4}{\text{ }}\]is mixed with 1 N KOH, the heat liberated is …….
A.11.45 KJ
B.57.3 KJ
C.573 J
D.573 KJ
Answer
498.6k+ views
Hint: A neutralisation reaction is a chemical reaction that occurs where an acid and a base react quantitatively to produce salt and water as products. In a neutralisation reaction, a mixture of H⁺ and $O{H^ - }$ ions combine to form vapour. An acid-base neutralisation reaction is the most common form of neutralisation reaction.
Complete answer:
Volumetric analysis is a commonly used quantitative analytical tool. As the name implies, this approach entails determining the concentration of an analyte by measuring the volume of a solution whose concentration is defined.
An uncertain number of chemicals are present in the solution to be examined.
To indicate the end-point, a reagent of uncertain concentration interacts with a chemical of unknown quantity in the presence of an indicator (usually phenolphthalein). It's the stage at which the reaction is over. Titration, which ends the reaction between the solution and the reagent, is used to quantify the quantities. The sum of reagent and solution is seen by the volume and concentration of reagent used in the titration. The mole fraction of the equation determines the quantity of unknown chemical in the same volume of solution.
Here \[{H_2}S{O_4}{\text{ }}\] is the limiting reactant
Using volumetric analysis
\[{N_1}{V_1} = {N_2}{V_2}\]
\[0.2 \times 50 = 1 \times {V_2}\]
\[{V_2} = \dfrac{2}{{10}} \times 50 = 10\;{\text{mL}}\]
Thus 50ml of \[0.2{\text{N}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] will neutralise exactly with 10 ml of 1N KOH
Heat produced on neutralisation of 1000 ml of 1N KOH = 57.3 KJ
Heat produced in neutralisation of10mL of 1N KOH \[ = \dfrac{{57.3}}{{1000}} \times 10 \times 1000J = 573J\]
Option C is correct.
Note:
The transition in enthalpy that happens when one equivalent of an acid and a base undergo a neutralisation reaction to form water and a salt is known as the enthalpy of neutralisation. It is a special case of reaction enthalpy. It is known as the amount of energy released when one mole of water is formed.
Complete answer:
Volumetric analysis is a commonly used quantitative analytical tool. As the name implies, this approach entails determining the concentration of an analyte by measuring the volume of a solution whose concentration is defined.
An uncertain number of chemicals are present in the solution to be examined.
To indicate the end-point, a reagent of uncertain concentration interacts with a chemical of unknown quantity in the presence of an indicator (usually phenolphthalein). It's the stage at which the reaction is over. Titration, which ends the reaction between the solution and the reagent, is used to quantify the quantities. The sum of reagent and solution is seen by the volume and concentration of reagent used in the titration. The mole fraction of the equation determines the quantity of unknown chemical in the same volume of solution.
Here \[{H_2}S{O_4}{\text{ }}\] is the limiting reactant
Using volumetric analysis
\[{N_1}{V_1} = {N_2}{V_2}\]
\[0.2 \times 50 = 1 \times {V_2}\]
\[{V_2} = \dfrac{2}{{10}} \times 50 = 10\;{\text{mL}}\]
Thus 50ml of \[0.2{\text{N}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] will neutralise exactly with 10 ml of 1N KOH
Heat produced on neutralisation of 1000 ml of 1N KOH = 57.3 KJ
Heat produced in neutralisation of10mL of 1N KOH \[ = \dfrac{{57.3}}{{1000}} \times 10 \times 1000J = 573J\]
Option C is correct.
Note:
The transition in enthalpy that happens when one equivalent of an acid and a base undergo a neutralisation reaction to form water and a salt is known as the enthalpy of neutralisation. It is a special case of reaction enthalpy. It is known as the amount of energy released when one mole of water is formed.
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