Δ H and ΔS for a reaction are +30.558 kJ mol-1 and 0.066 kJ K-1mol-1 at 1 atm pressure. The temperature at which free energy change is equal to zero and the nature of the reaction below this temperature are :
483 K, spontaneous
443 K, non-spontaneous
443 K, spontaneous
463 K, non-spontaneous
D.
463 K, non-spontaneous
ΔG= ΔH- TΔS
where,ΔH = 30.558 kJK-1 mol-1
ΔS = 0.066 kJK' mol
ΔG = 0
0 = ΔH- TΔS
ΔH = TΔS
30.558 = T x 0.066
T = 463K
Reaction is non-spontaneous reason being ΔG = 0.
The enthalpy of vaporization of substance is 840 J mol-1 and its boiling point is -173°C. Its entropy of vaporization is :
42 J mol-1 K-1
21 J mol-1 K-1
84 J mol-1 K-1
8.4 J mol-1 K-1
Given the following thermochemical equations:
Zn + O2 → ZnO + 84,000 cal
Hg + O2 → HgO + 21,700 cal
Accordingly the heat of reaction for the following reaction, Zn + HgO → Hg + heat is:
105,700 cal
61,000 cal
62,300 cal
60,000 cal
The law of thermodynamics formulated by Dr.N. Nernst is :
first law of thermodynamics
second law of thermodynamics
third law of thermodynamics
both (a) and (b)
Heat of neutralization will be minimum for which of the following combination?
NaOH + H2SO4
NH4OH + CH3COOH
NaOH + HCl
NaOH + CH3COOH
2 moles of helium gas expanded isothermally and irreversible at 27°C from volume 1 dm3 to 1 m3 at constant pressure of 100 kPa. Calculate the work done.
99900 kJ
99900 J
34464.65 kJ
34464.65 J
Heat of formation of SO2 is -298 kJ. What is the heat of combustion of 4 g of S ?
+ 37 kJ
-37.15 kJ
+ 298 kJ
18.6 kJ
Enthalpy (H) is equal to
internal energy (E)
product of pressure (p) and volume (V) of gas
internal energy (E)+ pV
work (W) done by a system