A man takes a diet equivalent to 10000 kJ per day and does work, in expending his energy in all forms equivalent to 12500 kJ per day. What is change in internal energy per day?If the energy lost was stored as sucrose (1632 kJ pre 100g), how many days should it take to lose 2kg of his weight?(Ignore water loss)

The enthalpy of vaporization of liquid diethyl ether (C₂H₅)₂₀ is 26.0 KJ mol^-1 at its boiling point (35.0^oC). Calculate \(\triangle \)S^o for the conversion of Liquid to vapour and

Calculate the standard Gibbs energy change for the formation of propane at 298 K.
3 C (graphite) + 4H₂(g) → C₃H₈ (g)
\(\triangle _{ f }{ H }^{ o }for\quad propane,{ C }_{ 3 }{ H }_{ 8 }(g)=270.2J{ K }^{ -1 }{ mol }^{ -1 }\)
\({ S }_{ m }^{ o }C(graphite)=5.70J{ K }^{ -1 }{ mol }^{ -1 }\)
\( and \ { S }_{ m }^{ 0 }{ H }_{ 2 }(g)=130.7J{ K }^{ -1 }{ mol }^{ -1 }\)

The molar heat of formation of NH₄NO₃ (s) is -367.54 kJ and those of N₂ O(g) and H₂O(i) are + 81.46 kJ and -258.78 kJ respectively at 25^oC and 1.0 atmospheric pressure. Calculate \(\Delta \)H and \(\Delta \)U for the reaction.

Justify the following statements.
(i) Reaction with \(\triangle \)G^o < 0 always have an equilibrium constant greater than 1.
(ii) Many thermodynamically feasible reactions do not occur under ordinary conditions.
(iii) At low temperatures, enthalpy change dominates the △△G expression and at high temperatures, it is the entropy which dominates the value of △G.

Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
\( CH_{ 3 }OH(l)+\frac { 3 }{ 2 } O_{ 2 }(g)\rightarrow CO_{ 2 }(g)+2H_{ 2 }O(l);\Delta _{ r }{ H }^{ o }=-726 \ kJ \ { mol }^{ -1 }\)
\( C_{ (graphite) }+O_{ 2 }(g)\rightarrow CO_{ 2 }(g);\Delta _{ c }{ H }^{ o }=-393 \ kJ \ { mol }^{ -1 }\)
\( H_{ 2 }(g)+\frac { 1 }{ 2 } O_{ 2 }(g)\rightarrow H_{ 2 }O(l);\Delta _{ f }{ H }^{ o }=-286 \ kJ \ { mol }^{ -1 }\)

Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. ∆_fusH = 6.03 kJ mol^–1 at 0°C.
C_p [H2O(l)] = 75.3 J mol^–1 K^–1
C_p [H2O(s)] = 36.8 J mol^–1 K^–1 .

Calculate the enthalpy of the reaction:

When liquid benzene is oxidised at constant pressure at 300 K, the change in enthalpy is -3728 kJ. What is the change in internal energy at the same temperature?

The enthalpy of formation of methane at constant pressure and 300 K is - 78.84 kJ. What will be the enthalpy of formation at constant volume?

Calculate the enthalpy change for the reaction: H₂(g) +Cl₂(g)\(\longrightarrow \)2HCl(g). Given that bond energies of H - H, Cl- Cl and H - Cl bonds are 433, 244 and 431 kJ mol^-1 respectively.

Calculate the bond energy of C - H bond, given that the heat of formation of CH₄ heat of sublimation of carbon and heat of dissociation of H₂ are - 74.8, + 719.6, 435.4 kJ mol^-1 respectively.

give the reason for the following.
(a)Neither q nor w is a state function but q + w is a state function.
(b) A real crystal has more entropy than an ideal crystal.

Under what condition, the heat evolved or absorbed in a reaction is equal to its free energy change?
(b) Calculate the entropy change for thefollowing reversible process
\({ H }_{ 2 }O(s)\rightleftharpoons { H }_{ 2 }O(l){ \triangle }_{ fus }His \ 6.0 \ KJ \ { mol }^{ -1 }\)
\( \triangle G=\triangle H-T\triangle S\)