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
I am thinking it's D but i am still confused :S
It is not d.
Reverse equation 1 (and the sign of DH) and add to equation 2 and see if you don't get the sublimation equation.
N2O5(s) ==> N2O5(g)
Look this is what i did
You multiply the first reaction by -1
so you have -x for the reactant (since its what you started with)
and you have y for the product (since it makes a gas)
So enthalpy change is Delta H=>H(products)- H(reactants)
which means its y-(-x)
negatives cancel out
giving you y+x or x+y
Is that right? That was my first answer but now i'm not sure -_-
And btw thank you a lot for helping :)
No, that isn't right.
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 = -x kJ.
Then add equation 2.
N2 + 5/2 O2 ==> N2O5(g) DH = y kJ.
-----------------------------------
N2 and 5/2 O2 cancel leaving
N2O5(s) ==> N2O5(g) DH = sum of the above DH values which is -x+y
right?
So it's C
You MAY multiply by -1 as you did but then you ADD the two kJ values since you ADDED the two equations. Your error is the products - reactants thing. That doesn't apply here; that's how the original value of x and y were determined.
Yes, c is correct.
To determine the enthalpy change for the sublimation of one mole of N2O5 solid to gas, we need to use Hess's Law. According to Hess's Law, the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of all the individual reactions involved.
In this case, we have two thermochemical equations:
1) N2 + 2.5O2 -> N2O5(s) ∆H = x kJ
2) N2 + 2.5O2 -> N2O5(g) ∆H = y kJ
We want to find the enthalpy change for the sublimation of N2O5(s) to N2O5(g). We can do this by considering the difference between the two equations.
Sublimation is the phase transition from solid to gas without passing through the liquid phase. In this case, we can see that in both equations, N2O5 is formed from N2 and O2. Thus, if we subtract equation 1 from equation 2, we will cancel out N2 and O2 and be left with the desired sublimation reaction:
N2O5(g) - N2O5(s)
The enthalpy change for this sublimation reaction can be found by subtracting equation 1 from equation 2:
N2O5(g) - N2O5(s) ∆H = (N2 + 2.5O2 -> N2O5(g) ∆H) - (N2 + 2.5O2 -> N2O5(s) ∆H)
= y kJ - x kJ
= y - x kJ
Therefore, the correct answer is (c) y - x.