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Thermodynamics question

2024 · 9 Apr · Shift 1 · Q28
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Thermodynamics question

2024 · 9 Apr · Shift 1 · Q28

JEE MainChemistryThermodynamicsNumerical+4 / −1
When equal volume of 1 M HCl1 \mathrm{~M} \mathrm{~HCl}1 M HCl and 1 M H2SO41 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_41 M H2​SO4​ are separately neutralised by excess volume of 1MNaOH1 \mathrm{M}\mathrm{NaOH}1MNaOH solution. xxx and y kJy \mathrm{~kJ}y kJ of heat is liberated respectively. The value of y/xy / xy/x is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. Heat of neutralisation for strong acid–strong base

For a strong acid neutralised by a strong base, the enthalpy of neutralisation is approximately constant per mole of water formed:

ΔHneut≈−57 kJ mol−1\Delta H_{\text{neut}} \approx -57\, \text{kJ mol}^{-1}ΔHneut​≈−57kJ mol−1

So, the heat liberated is proportional to the number of moles of H+\mathrm{H^+}H+ neutralised, or equivalently, the moles of water formed.

  1. Case 1: 1 M1\,\mathrm{M}1M HCl

Let the equal volume taken in each case be VVV litre.

For 1 M1\,\mathrm{M}1M HCl:

moles of HCl=1×V=V\text{moles of HCl} = 1 \times V = Vmoles of HCl=1×V=V

Since HCl is monoprotic:

HCl+NaOH→NaCl+H2O\mathrm{HCl + NaOH \to NaCl + H_2O}HCl+NaOH→NaCl+H2​O

Thus, moles of water formed =V= V=V.

Hence heat liberated:

x=57V(in kJ, proportional form)x = 57V \quad (\text{in kJ, proportional form})x=57V(in kJ, proportional form)
  1. Case 2: 1 M1\,\mathrm{M}1M H_2SO_4$

For 1 M1\,\mathrm{M}1M H_2SO_4$:

moles of H2SO4=1×V=V\text{moles of } H_2SO_4 = 1 \times V = Vmoles of H2​SO4​=1×V=V

Sulfuric acid is diprotic:

H2SO4+2NaOH→Na2SO4+2H2O\mathrm{H_2SO_4 + 2NaOH \to Na_2SO_4 + 2H_2O}H2​SO4​+2NaOH→Na2​SO4​+2H2​O

Thus, moles of water formed =2V= 2V=2V.

Hence heat liberated:

y=57×2V=114Vy = 57 \times 2V = 114Vy=57×2V=114V
  1. Find the ratio y/xy/xy/x
yx=114V57V=2\frac{y}{x} = \frac{114V}{57V} = 2xy​=57V114V​=2
  1. Final answer
2\boxed{2}2​
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