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Consider the following two thermochemical equations N2+2.5O2-> N2O5(s) Delta H=xkJ N2+2.5O2-> N2O5(g) Delta H=ykJ The enthalpy...Asked by Bobz
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
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
Answers
Answered by
DrBob222
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)
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)
Answered by
Bobz
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 :)
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 :)
Answered by
DrBob222
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?
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?
Answered by
Bobz
So it's C
Answered by
DrBob222
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.
Answered by
DrBob222
Yes, c is correct.
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