Advertisements
Advertisements
Question
The standard molar entropy of \[\ce{H2O (l)}\] is 70 JK–1 mol–1. Will the standard molar entropy of \[\ce{H2O (s)}\] be more, or less than 70 JK–1 mol–1?
Solution
The standard molar entropy of \[\ce{H2O (l)}\] is 70 JK–1 mol–1. The solid form of \[\ce{H2O}\] is ice. In ice, molecules of \[\ce{H2O}\] are less random than in liquid water. Thus, molar entropy of \[\ce{H2O (s)}\] < molar entropy of \[\ce{H2O (l)}\]. The standard molar entropy of \[\ce{H2O (s)}\] is less than 70 JK–1 mol–1.
APPEARS IN
RELATED QUESTIONS
Heat of combustion is always ___________.
Molar heat of vapourisation of a liquid is 4.8 kJ mol–1. If the entropy change is 16 J mol–1 K–1, the boiling point of the liquid is
Define molar heat capacity.
Write the unit of molar heat capacity.
During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is ______.
\[\ce{∆_f U^Θ}\] of formation of \[\ce{CH4 (g)}\] at certain temperature is – 393 kJ mol–1. The value of \[\ce{∆_f H^Θ}\] is ______.
On the basis of thermochemical equations (a), (b) and (c), find out which of the algebric relationships given in options (i) to (iv) is correct.
(a) \[\ce{C (graphite) + O2 (g) -> CO2 (g) ; ∆_rH = xkJ mol^{-1}}\]
(b) \[\ce{C (graphite) + 1/2 O2 (g) -> CO (g) ; ∆_rH = ykJ mol^{-1}}\]
(c) \[\ce{CO (g) + 1/2 O2 (g) -> CO2 (g) ; ∆_r H = zkJ mol^{-1}}\]
If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
Derive the relationship between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.
The molar heat of formation of NH4NO3 (s) is −367.54 kJ and those of N2O (g), H2O (l) are 81.46 and −285.8 kJ respectively at 25°C and atmosphere pressure. The difference of ΔH and ΔE of the reaction \[\ce{NH4NO3(s) -> N2O (g) + 2H2O (l)}\] is ______ kJ.
