Advertisements
Advertisements
प्रश्न
Enthalpy of neutralization is always a constant when a strong acid is neutralized by a strong base: account for the statement.
Advertisements
उत्तर
- Enthalpy of neutralization of a strong acid by a strong base is always a constant and it is equal to − 57.32 kJ irrespective of which acid or base is used.
- Because strong acid or strong base means it is completely ionized in solution state. For e.g., NaOH (strong base) is neutralized by HCl (strong acid), due to their complete ionization, the net reaction takes place in only water formation.
So the enthalpy of neutralization is always constant for strong acid by a strong base.
\[\ce{H^+Cl^- + Na^+OH^- -> Na^+Cl^- + H2O}\]
\[\ce{H^+NO3^+ + K^+OH^- -> K^+NO3^+ + H2O}\]
(Net reaction) \[\ce{H^+ + OH^- -> H2O}\]
∆H = − 57.32 kJ
APPEARS IN
संबंधित प्रश्न
Calculate the amount of work done in the
1) Oxidation of 1 mole HCl(g) at 200 °C according to reaction.
4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)
2) Decomposition of one mole of NO at 300 °C for the reaction
2NO(g) → N2(g) + O2(g)
Calculate the work done and comment on whether work is done on or by the system for the decomposition of 2 moles of NH4NO3 at 100 °C
NH4NO3(s) → N2O(g) + 2H2O(g)
Define enthalpy of combustion.
What is standard N ≡ N bond enthalpy from following reaction,
\[\ce{N2_{(g)} + 3H2_{(g)} -> 2NH3_{(g)}; \Delta H^0 = - 83 kJ}\]
\[\ce{ΔH^0_{(H-H)}}\] = 435 kJ; \[\ce{ΔH^0_{(N-H)}}\] = 389 kJ
The difference between heats of reaction at constant pressure and at constanl volume for the reaction
\[\ce{2C6H6_{(l)} + 15O2_{(g)} -> 12CO2_{(g)} + 6H2O_{(l)}}\] at 25°C in kJ
In which of the following reactions does the heat change represent the heat of formation of water?
In which of the following reactions, ∆H is greater than ∆U?
For the reaction, \[\ce{N_{2(g)} + 3H_{2(g)} -> 2NH_{3(g)}}\], ΔH is equal to ______.
Calculate the work done in oxidation of so2(g) at 25°C if, \[\ce{2SO_{2(g)} + O2_{(g)} -> 2SO_{3(g)}}\], R = 8.314 J K−1 mol−1.
Define enthalpy.
