QJEEM24Ch0e8a: JEE Main 2024 Chemistry - Class 11 JEEM24Ch0e8a Solution: When equal volume of 1   M   HCl and 1   M…

To solve this problem, we need to understand the concept of neutralization reactions and the heat evolved during these reactions.

When an acid and a base react, they undergo a neutralization reaction to produce water and a salt. The heat released in this process is known as the enthalpy of neutralization.

Consider the neutralization of hydrochloric acid (HCl) and sulfuric acid (H₂SO₄) by sodium hydroxide (NaOH). The balanced chemical equations for these neutralizations are:

$$\mathrm{HCl}+\mathrm{NaOH}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$$

$${\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$$

For the first reaction, each mole of HCl reacts with one mole of NaOH, releasing a certain amount of heat (let's denote this amount by $x$ kJ). For the second reaction, each mole of H₂SO₄ reacts with two moles of NaOH. Since we are given equal volumes and molarities of HCl and H₂SO₄, we can infer that one mole of H₂SO₄ will produce twice the heat of one mole of HCl because it produces double the amount of water.

Thus, the heat evolved in the neutralization of H₂SO₄ by NaOH (denoted as $y$ kJ) will be approximately twice that of HCl. Therefore, $y = 2x$ .

Therefore, the value of $$\frac{y}{x}$$ is:

$$\frac{y}{x}=\frac{2x}{x}=2$$

So, the value of $$\frac{y}{x}$$ is 2.