Consider the following two thermochemical equations
N2+2.5O2-> N2O5(s) Delta H=xkJ
N2+2.5O2-> N2O5(g) Delta H=ykJ
The enthalpy change in kJ for the sublimation of one mole of N205 solid to gas would be represented by the quantity
a)x+y
b)x-y
c)y-x
d)-x-y
Assuming it's a)
No. Sublimation is changing from solid to gas. Reverse equation 1 and add to equation 2. You should get
N2O5(s) ==> N2O5(g), then reverse kJ for equation 1 and add to equation 2.
So the answer is a?
I don't think so.
N2+2.5O2-> N2O5(s) Delta H=xkJ
N2+2.5O2-> N2O5(g) Delta H=ykJ
Reverse equation 1.
N2O5(s) ==> N2 + 5/2 O2 DH = -xkJ
N2 + 5/2 O2 ==> N2O5(g) DH = +ykJ.
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N2O5(s) ==>N2O5(g) DH = -x+y
To determine the enthalpy change for the sublimation of one mole of N2O5 from the solid state to the gas state, we can use the concept of Hess's Law. This law states that if a reaction can be expressed as the sum of two or more other reactions, then the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.
In this case, we have two thermochemical equations:
1) N2 + 2.5O2 -> N2O5(s) with ΔH = x kJ
2) N2 + 2.5O2 -> N2O5(g) with ΔH = y kJ
We want to find the enthalpy change for the sublimation of N2O5(s) to N2O5(g), which can be represented as the sum of the following reactions:
1) N2O5(s) -> N2O5(g)
2) N2 + 2.5O2 -> N2O5(g) (reverse of the second equation)
According to Hess's Law, the enthalpy change for the overall reaction is equal to the sum of the enthalpy changes of the individual reactions:
ΔH(sublimation) = ΔH1 + ΔH2(reverse)
= ΔH1 - ΔH2
= x - y kJ
Therefore, the correct quantity that represents the enthalpy change for the sublimation of one mole of N2O5 solid to gas is (a) x - y.