ThermodynamicsHard

Question

One mole of ice is converted into water at 273 K and 1 atm. The entropies of H2O (s) and H2O (l) are 38.0 and 58.0 J/K-mol, respectively. The enthalpy change for the conversion is

Options

A.5460 kJ/mol
B.5460 J/mol
C.−5460 J/mol
D.20 J/mol

Solution

$\Delta S = S_{\text{water}} - S_{\text{ice}} = 58.0 - 38.0 = 20\text{ J/K.mol}$

Now, $\Delta H = T.\Delta S = 273 \times 20 = 5460\text{ J/mol}$

Create a free account to view solution

View Solution Free
Topic: Thermodynamics·Practice all Thermodynamics questions

More Thermodynamics Questions

The magnitude of work done by one mole of a Van der Waals gas during its isothermal reversible expansion from volume V1 ...For which of the following ideal gas, Cv,m is independent of temperature?...An ideal gas with the adiabatic exponent $\gamma$goes through a process P = Po – αV, where Po and α are positive constan...One mole of oxygen is heated from 0°C at constant pressure till its volume is increased by 10%. The specific heat of oxy...For a reaction, A ⇌ P, the plots of [A] and [P] with time at temperatures T1 and T2 are given below. If T2 > T1...