Advertisements
Advertisements
Question
You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below.
| Liquid | Boiling points (°C) | ΔH (kJ mol−1) |
| Ethanol | 78.4 | + 42.4 |
Advertisements
Solution
Given, Tb = 78.4°C = (78.4 + 273) = 351.4 K
ΔHV(ethanol) = + 42.4 kJ mol−1
∆SV = `(Δ"H"_"V")/"T"_"b"`
∆SV = `(+ 42.4 "kJ mol"^-1)/(351.4 "K")`
∆SV = `(+ 42400 "J mol"^-1)/(351.4 "K")`
∆SV = + 120.66 J K−1 mol−1
APPEARS IN
RELATED QUESTIONS
The correct thermodynamic conditions for the spontaneous reaction at all temperature is
The values of ∆H and ∆S for a reaction are respectively 30 kJ mol–1 and 100 JK–1 mol–1. Then the temperature above which the reaction will become spontaneous is
Identify the state and path functions out of the following:
- Enthalpy
- Entropy
- Heat
- Temperature
- Work
- Free energy
State the various statements of the second law of thermodynamics.
What are spontaneous reactions?
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK−1 mol−1. Calculate the melting point of sodium chloride.
You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below.
| Liquid | Boiling points (°C) | ΔH (kJ mol−1) |
| Toluene | 110.6 | + 35.2 |
Cyanamide (NH2CN) is completely burnt in excess oxygen in a bomb calorimeter, ΔU was found to be −742.4 kJ mol−1, calculate the enthalpy change of the reaction at 298 K.\[\ce{NH2CN_{(s)} + 3/2 O2_{(g)} -> N2_{(g)} + CO2_{(g)} + H2O_{(l)}}\] ΔH = ?
Calculate the enthalpy of hydrogenation of ethylene from the following data.
Bond energies of C − H, C − C, C = C and H − H are 414, 347, 618 and 435 kJ mol−1.
For the reaction at 298 K: \[\ce{2A + B -> C}\]
ΔH = 400 KJ mol−1; ΔS = 0.2 KJ K−1 mol−1 Determine the temperature at which the reaction would be spontaneous.
