The maximum number of molecule is present in
15 L of H2 gas at STP
5 L of N2 gas at STP
0.5 g of H2 gas
10 g of O2 gas
The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm, is
-6 J
-608 J
+304 J
-304 J
Born-Haber cycle may be used to calculate
electronegativity
mass number
oxidation number
electron affinity
The heat of formation for CO2 (g), H2O (l) and CH4 (g) are -400 kJ mol-1, -280 kJ mol-1 and -70 kJ mol-1 respectively. The heat of combustion of CH4 in kJ mol-1 is
890
-160
-890
-90
C (s) + O2 (g) → CO2 (g); H = -94kcal
2CO (g) + O2 (g) → 2CO2 (g); H = -135 kcal.
The heat of formation of CO (g) is
-26.4 kcal
41.2 kcal
26.4 kcal
229.2 kcal
The entropy of a perfectly crystalline material is zero at 0°C. This is statement of
first law of thermodynamics
second law of thermodynamics
third law of thermodynamics
law of conservation of energy
In any chemical reaction, a quantity that decrease to a minimum is
free energy
entropy
temperature
enthalpy
Given, H2O (l) → H2O (g); H1 = +43 kJ
H2O (s) → H2O (l); H2 = +6.05 kJ. Calculate the enthalpy of sublimation of ice.
49.75 kJ mol-1
37.65 kJ mol-1
43.7 kJ mol-1
55.23 kJ mol-1
If standard enthalpies of formation of CaCl (s) (hypothetical) and that of CaCl2 (s) are -188 J mol-1 and -795 kJ mol-1 respectively, calculate the value of standard heat of reaction for the following disproportionation reaction
2CaCl (s) → CaCl2 (s) + Ca (s)
-607 kJ mol-1
+607 kJ mol-1
-419 kJ mol-1
+419 kJ mol-1
Heat formation of H2O is -188 kJ/ mol and H2O2 is 286 kJ/mol. The enthalpy change for the reaction; 2H2O2 → 2H2O + O2 is
196 kJ
-196 kJ
984 kJ
-984 kJ