For which one of the following reactions will the enthalpy change be approximately equal to the internal energy change?
2 NaN3(s) → 2 Na(s) + 3 N2(g)
2 H2O( ℓ ) → 2 H2(g) + O2(g)
2 Sb(s) + 3 I2(g) → 2 SbI3(s)
Ca(NO3)2(aq) + 2 NH4Cl(aq) → CaCl2(aq) + 2 NH4NO3(aq)
CaCO3(s) → CaO(s) + CO2(g)
As we know that ,
∆H= ∆U +P∆V
Or
∆H = ∆U + ∆ngRT
Where , ∆H = enthalpy change
∆U = internal energy change
∆ng = change in no. of moles of gas= no. of gaseous moles in product side - no. of gaseous moles in reactant side
R = gas constant
T = temperature
Now if a reaction has ∆ng = 0 , means it has ∆nRT ( work done) zero or in other words we can say that if ∆ng = 0 then enthalpy change will be equals to internal energy change.
Now for reaction 1 ,
2NaN3 (s) --> 2Na(s) + 3N2(g)
we see that ∆ng = 3 ,thus ∆ngRT will not be zero it means its enthalpy change will be not equals to internal energy change.
For reaction 2 ,
2H2O(l) --> 2H2(g) +O2(g)
again ∆ng = 3 which again indicates that enthalpy change is not equals to internal energy change.
For reaction 3 ,
2Sb(s) +3I2(g)--> 2SbI3(s)
We see that ∆ng = -3 it means it's enthalpy change will be not equals to internal energy change.
For reaction 4 ,
Ca(NO3)2(aq) + 2NH4Cl(aq) --> CaCl2(aq) +2NH4NO3(aq)
we see that ∆ng = 0 , it means it's enthalpy change will be equals to internal energy change.moreover since all species are in aqueous state thus ∆V will also be zero , hence P∆V will be zero , thus enthalpy is approximately equal to change in internal energy
For reaction 5
CaCO3(s) --> CaO(s) + CO2(g)
, we see that ∆ng = 1 which indocates that enthalpy change is not equals to internal energy change.
Thus the correct answer is reaction 4 that is
Ca(NO3)2(aq) + 2NH4Cl(aq) --> CaCl2(aq) + 2NH4NO3(aq)
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