Advertisements
Advertisements
प्रश्न
Define enthalpy of neutralization.
Advertisements
उत्तर
The enthalpy of neutralization is defined as the change in enthalpy of the system when one gram equivalent of an acid is neutralized by one gram equivalent of a base or vice versa in dilute solution.
\[\ce{H^+_{(aq)} + OH^-_{(aq)} -> H2O_{(l)}}\] = 57.32 kJ
APPEARS IN
संबंधित प्रश्न
Obtain the relationship between ΔH and ΔU for gas phase reactions.
Calculate the standard enthalpy of formation of \[\ce{CH3OH_{(l)}}\] from the following data:
\[\ce{CH3OH_{(l)} + 3/2 O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)} }\]; ΔrH° = − 726 kJ mol-1
\[\ce{C_{(graphite)} + O2_{(g)} -> CO2_{(g)}}\]; ΔcH° = −393 kJ mol−1
\[\ce{H2_{(g)} + 1/2 O_{(g)} -> H2O_{(l)}}\]; ΔfH° = −286 kJ mol−1
An ideal gas expands from the volume of 1 × 10–3 m3 to 1 × 10–2 m3 at 300 K against a constant pressure at 1 × 105 Nm–2. The work done is
Enthalpy of neutralization is always a constant when a strong acid is neutralized by a strong base: account for the statement.
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
Identify the equation in which change in enthalpy is equal to change in internal energy.
The work done during combustion of 9 × 10-2 kg of ethane, C2H6 (g) at 300 K is ______.
(Given R = 8.314 J deg-1, atomic mass C = 12, H = 1)
Define enthalpy.
The equilibrium concentrations of the species in the reaction \[ \mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D} \] are 2, 3, 10 and 6 mol L-1, respectively at 300 K. ΔGº for the reaction is (R = 2 cal/mol K)
