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 __________.

<p>To solve this problem, we need to understand the concept of neutralization reactions and the heat evolved during these reactions. <p>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.</p> <p>Consider the neutralization of hydrochloric acid (HCl) and sulfuric acid (H₂SO₄) by sodium hydroxide (NaOH). The balanced chemical equations for these neutralizations are:</p> <p>$\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}$</p> <p>$\mathrm{H_2SO_4} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2SO_4} + 2\mathrm{H_2O}$</p> <p>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. </p> <p>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 $. <p> <p>Therefore, the value of $\frac{y}{x}$ is:</p> </p> <p>$\frac{y}{x} = \frac{2x}{x} = 2$</p> <p>So, the value of $\frac{y}{x}$ is 2.</p> </p>