When equal volume of $$1$$ M HCl and $$1$$ M $$H_2SO_4$$ are separately neutralised by excess volume of $$1$$ M NaOH solution. $$x$$ and $$y$$ kJ of heat is liberated respectively. The value of $$y/x$$ is _______

When 1 M HCl (strong monoprotic acid) is neutralized by excess NaOH:

$$HCl + NaOH \rightarrow NaCl + H_2O$$

Each mole of HCl releases 1 mole of $$H^+$$. For volume V: moles of $$H^+$$ = V × 1 = V mol.

Heat released = $$x$$ kJ.

When 1 M $$H_2SO_4$$ (strong diprotic acid) of the same volume is neutralized:

$$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$

Each mole of $$H_2SO_4$$ releases 2 moles of $$H^+$$. For volume V: moles of $$H^+$$ = V × 2 = 2V mol.

Heat released = $$y$$ kJ.

Since the heat of neutralization is per mole of $$H^+$$ (for strong acid-strong base), and $$H_2SO_4$$ produces twice as many $$H^+$$ ions as HCl for equal volumes at equal molarity:

$$y = 2x$$

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

The answer is $$\boxed{2}$$.