Difference between Internal Energy and Enthalpy
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Q. Internal energy change for combustion of ethanol is 416.5 kJ mol−1. The calorific value of ethanol at 300 K in (kJ) is (Take R= 25/3 J/molK)
Q. If ΔH is the change in enthalpy and ΔE the change in internal energy accompanying a gaseous reaction
- ΔH<ΔE only if the number of moles of the products is greater than the number of the reactants
- ΔH is always greater than ΔE
- ΔH is always less than ΔE
- ΔH<ΔE only if the number of moles of the products is less than the number of moles of the reactants
Q. The difference between heats of reaction at constant pressure and at constant volume for the reaction 2C6H6(l)+15 O2(g)→12CO2(g)+6H2O(l) at 25∘C at in kJ is
- + 7.43
- –3.72
- + 3.72
- –7.43
Q. Which is not the correct relation between enthalpy ΔH and intrinsic energy ΔE
- ΔH=ΔE−PΔV
- ΔH=ΔE+PΔV
- ΔH=ΔE+nRT
- ΔE=ΔH−PΔV
Q. For the reaction N2+3H2⇌2NH3;ΔH=
- ΔE−RT
- ΔE−2RT
- ΔE+2RT
- ΔE+RT
Q. At constant T and P, which one of the following statements is correct for the reaction, CO(g)+12O2(g)→CO2(g)
- ΔH is independent of the physical state of the reactants of that compound
- ΔH>ΔE
- ΔH<ΔE
- ΔH=ΔE
Q. The enthalpy of formation of methane at constant pressure and 300 K is – 78.84 KJ. The enthalpy of formation at constant volume is:
- – 75.5 kJ
- + 75.5 kJ
- – 102.6 kJ
- – 54.0 kJ
Q.
The enthalpy change for the reaction of 50.00 ml of ethylene with 50.00 ml of H2 at 1.5 atm pressure is ΔH− = -0.31 kJ. The value of ΔE will be
2.567 kJ
- 0.3024 kJ
0.3024 kJ
-0.0076 kJ
Q. What is the relation between ΔH and ΔE in this reaction?
CH4(g)+2O2(g)⇌CO2(g)+2H2O(l)
CH4(g)+2O2(g)⇌CO2(g)+2H2O(l)
Q. For the reaction, C(s)+O2(g)→CO2(g)
- Cannot be predicted
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