The internal energy change
(∆U) for a reaction can be calculated from the enthalpy change
(∆H) by using the relationship:
∆U=∆H−∆(pV)In instances where the process occurs at constant pressure, we can simplify the term
∆(pV) to
∆(nRT). The change in the number of moles of gas
(∆n) and the relation to the decrease in volume at constant pressure and temperature provide the basis for this relationship:
∆(pV)=∆(nRT)=(nproduct −nreactant )(RT)Given that Nitrous oxide
(N2O) decomposes into Nitrogen
(N2) and Oxygen
(O2), represented by the equation:
2N2O(g)⟶2N2(g)+O2(g)We now calculate the change in moles of gas
(∆n) as follows:
Moles of gas initially (from
N2O):2 moles
Moles of gas produced (from
N2 and
O2 ):
2+1=3 moles
The change in moles
(∆n) is:
∆n=3−2=1 mole Using the ideal gas law, where
R=8.314J∕mol⋅K and temperature
T=298K, we calculate:
∆(nRT)=(1 mole )×(8.314J∕mol⋅K)×(298K) ∆(nRT)=2476.372J=2.476kJNow, using the given value of
∆H=−163.15kJ for the decomposition of 2 moles of
N2O, we apply it to calculate
∆U :
∆U=∆H−∆(nRT)∆U=−163.15kJ−2.476kJ=−165.626kJTo find the
∆U per
100g of
N2O, we need the molar mass of
N2 which is
44.013g∕mol :
Total moles in
100g of
N2O :
Moles
==2.272 moles
The
∆U per
100g of
N2O will then be proportional:
∆U100g=−165.626kJ×=−188.235kJThis calculated internal energy change for
100g of Nitrous oxide closely rounds to: