Enthalpy
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- ΔU=0, ΔStotal≠0
- ΔU=0, ΔStotal=0
- ΔU=0, ΔStotal=0
- ΔU≠0, ΔStotal≠0
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0∘C to ice at −10.0∘C.Δ fusH=6.03kJ mol−1 at 0∘C.
(Cp[H2O(l)]=75.3 J mol−1K−1
Cp[H2O(s)]=36.8 J mol−1K−1
- -5.763
- 5.763
- 1.013
- -1.013
A solution of two components containing moles of the component and moles of the second component is prepared. and are the molecular weights of component and respectively. If is the density of the solution in , is the molarity and is the mole fraction of the component, then can be expressed as:
Polymerisation of ethene to poly - ethene is represented by the equation
n(CH2=CH2)→(−CH2−CH2−)n
Given that average enthalpies of C = C and C - C bonds at 298 K are 590 and 331 kJ mol−1 respectively, predict the enthalpy change when 56g of ethene changes to polyethylene.
144 kJ
-144 kJ
72 kJ
-72 kJ
Liquid⇌Vapour
Which of the following relations is correct?
- d In PdT2=−ΔHvT2
- d In PdT=−ΔHvRT
- d In GdT2=ΔHvRT2
- d In PdT=ΔHvRT2
- 13 kJ
- 62 kJ
- 16 kJ
- 21 kJ
Define bond enthalpy and bond dissociation enthalpy
Enthalpy of sublimation for Ca(s) → Ca(g) = 121 kJ mol−1
Enthalpy of dissociation of Cl2(g) → 2Cl(g) =242.8 kJ mol−1
Ionisation energy of
Ca(g)→ Ca++(g) =2422 kJ mol−1
Electron gain enthalpy of 2Cl→ 2Cl−1=(2×−355) kJ mol−1=−710 kJ mol−1
Enthalpy of formation of CaCl2=−795 kJ mol−1
- −2670.8 kJ mol−1
- −2870.8 kJ mol−1
- −2770.8 kJ mol−1
- −2970.8 kJ mol−1
ΔH=−104 kJ/mol of P4
The enthalpy of sublimation [P4(s)→P4(g)] white is 59 kJ/mol and enthalpy of atomization is 316.25 kJ/mol of P(g).
Now find the average P−P bond enthalpy in P4 molecule is:
- 102 kJ
- 201 kJ
- 104 kJ
- 120 kJ
- ΔH=ΔE−P ΔV
- ΔH=ΔE+P ΔV
- ΔE=ΔV+ΔH
- ΔE=ΔH+PΔH
(Nearest integer)
(Given : R=8.3 J K−1mol−1)
The enthalpies of all elements in their standard states are?
Standard Molar Enthalpy of Formation of Is equal to
Calculate enthalpy change for the change 8S(g)→S8(g)′ given that
H2S2(g)→2H(g)+2S(g), ΔH=239.0 k cal mol−1H2S2(g)→2H(g)+S(g), ΔH=175.0 k cal mol−1
-508.0 k cal
- 512.0 k cal
508.0k cal
+ 512.0 k cal
H2(g)+12O2(g)→H2O(l), △H=−286 kJ
H2(g)+12O2(g), →H2O(g), △H=−245.5 kJ
- 1.25 kJ
- 40.5 kJ
- 6.02 kJ
- 62.3 kJ
(The molar heat of oxygen at constant pressure, Cp=7.03 cal mol−1 K−1 and R=8.31 J mol−1K−1)
- 352 cal
- 252 cal
- 452 cal
- 152 cal
250 K
400 K
450 K
600 K
When we add ice to water it becomes cold. Explain.
C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)
- 10.48 kJ
- 3.74 kJ
- 14.86 kJ
- 6.73 kJ
An ideal monoatomic gas of two moles occupying a volume V at.Gas expanding adiabatically having volume calculate (a)the final temperature of the gas and (b) change in its internal energy.
(a) K (b) kJ
(a) K (b) kJ
(a) K (b) kJ
(a) K (b) kJ
Explain what happens to the molecular motion and energy of of water at when it changes to the ice at the same temperature. How is the latent heat of fusion related to the energy exchange that takes place during this change of state?
Sublimation of lithium is 155.3 kJ mol−1
Dissociation of half mole of F2=75.3 kJ
Ionization enthalpy of lithium =520 kJ mol−1
Electron gain enthalpy of 1 mol of F(g) =−333 kJΔfHoverall=−594 kJ mol−1
- −1200.4 kJ mol−1
- −1488.6 kJ mol−1
- −1011.6 kJ mol−1
- −984.6 kJ mol−1
Assertion: Hydrogen gas has the highest calorific value.
Reason: Hydrogen is the lightest gas and hence produces a large amount of heat.
Both ‘A’ and ‘R’ are correct and ‘R’ is the correct explanation of ‘A’.
Both ‘A’ and ‘R’ are correct, but ‘R’ is not the correct explanation of ‘A’.
‘A’ is correct, but ‘R’ is incorrect
‘A’ is incorrect, but ‘R’ is correct
The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound
(a) is always negative
(b) is always positive
(c) may be positive or negative
(d) is never negative
ΔlatticeH=+788 kJmol−1
ΔsolH=+4 kJmol−1
- −784 kJmol−1
- −1084 kJmol−1
- +784 kJmol−1
- +1084 kJmol−1
- 7.9 km
- 9.7 km
- 4.8 km
- 8.4 km
C10H8 (s)+12O2 (g)→10CO2 (g)+4H2O (l)
at constant volume is given as −1228.2 kcal at 25oC. The heat of the reaction at constant pressure and same temperature is:
- −1345.5 kcal
- −1229.3 kcal
- −1156 kcal
- −1380.2 kcal
- 373 K
- 100∘C
- 0∘C
- −273∘C