Calculating ∆H for a Chemical Reaction

[courtrigrad]

Hello all:

Given the following data:

H2(g) + 1/2 O2(g) --> H20 (l) ∆H = -285.8 kJ

N2O5(g) + H20(l) --> 2HNO3(l) ∆H = -76.6 kJ

1/2 N2(g) + 3/2O2(g) + 1/2 H2 (g) --> HNO3(l) ∆ = -174.1 kJ

Calculate the ∆H for the reaction

2N2(g) + 5O2(g) --> 2N2O5(g)

My Solution

from N2: 2 = 1/2 *k3
from O2: 5 = 1/2 * k1 + 3/2 * k3
from N2O5: -2 = -k2

Hence ∆comb = ∆H1 * k1 + ∆H2 * k2 + ∆H3 * k3. I am not sure if this is right. Can someone please see if I made a mistake in my steps?

Thanks a lot

 

[apchemstudent]

Hello all:

Given the following data:

H2(g) + 1/2 O2(g) --> H20 (l) ∆H = -285.8 kJ

N2O5(g) + H20(l) --> 2HNO3(l) ∆H = -76.6 kJ

1/2 N2(g) + 3/2O2(g) + 1/2 H2 (g) --> HNO3(l) ∆ = -174.1 kJ

Calculate the ∆H for the reaction

2N2(g) + 5O2(g) --> 2N2O5(g)

My Solution

from N2: 2 = 1/2 *k3
from O2: 5 = 1/2 * k1 + 3/2 * k3
from N2O5: -2 = -k2

Hence ∆comb = ∆H1 * k1 + ∆H2 * k2 + ∆H3 * k3. I am not sure if this is right. Can someone please see if I made a mistake in my steps?

Thanks a lot

everything seems right...

 

[courtrigrad]

Why is the answer 28.4 kJ when I get -201.8?

 

[apchemstudent]

Why is the answer 28.4 kJ when I get -201.8?

do you have the correct signs? Can you show all your work...

 

[courtrigrad]

from N2: 2 = 1/2 * k3 k3 = 4
from O2: 5 = 1/2*k1 + 6. 1/2*k1 = -1, k1 = -2
from N2O5: -2 = -k2, k2 = 1

∆Comb = -2(-285.8) + 1( -76.6) + 4(-174.1)

 

[apchemstudent]

from N2: 2 = 1/2 * k3 k3 = 4
from O2: 5 = 1/2*k1 + 6. 1/2*k1 = -1, k1 = -2
from N2O5: -2 = -k2, k2 = 1

∆Comb = -2(-285.8) + 1( -76.6) + 4(-174.1)

k2 does not equal one

in fact

should be - k2 = 2 ( my bad... )

so k2 = -2..

 

[Gokul43201]

yes, that's the error : k2 = -2

Besides relying on the math, make sure the values you get make logical sense. Equation #2 has the N2O5 on the wrong side, so k2 must be -ve.

 

[courtrigrad]

thanks a lot guys!