JEE MAIN - Chemistry (2024 - 9th April Morning Shift - No. 28)

When equal volume of $$1 \mathrm{~M} \mathrm{~HCl}$$ and $$1 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4$$ are separately neutralised by excess volume of $$1 \mathrm{M}$$ $$\mathrm{NaOH}$$ solution. $$x$$ and $$y \mathrm{~kJ}$$ of heat is liberated respectively. The value of $$y / x$$ is __________.
Answer
2

Explanation

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:

$$ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} $$

$$ \mathrm{H_2SO_4} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2SO_4} + 2\mathrm{H_2O} $$

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.

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