Calculate DeltaH between 700K and 800K

[il postino]

Homework Statement

For the reaction 2NO2 <---> 2NO + O2
It is known to reach equilibrium at a total pressure of 1 atm.
And it is known that the ratio pNO / pNO2 = 0.872 at 700K
and the ratio pNO / pNO2 = 2.5 at 800K.
Calculate the enthalpy of reaction between 700 and 800 K

Relevant Equations

Van´t Hoff

Hi all
Knowing that $kp$ is:
$k_p=\frac{pO_2.(pNO)^2}{(pNO_2)^2}$
And knowing these relationships between the partial pressures, I obtained:
At 700 K:
$pNO = 0.872.pNO2$
$k_p=pO_2.(0.872)^2$

At 800 K:
$pNO = 2.5.pNO2$
$k_p=pO_2.(2.5)^2$

Furthermore it is known that $pNO + pNO + PO2 = 1 atm$

We also know that:
$\Delta H=\frac{-ln(\frac{kp2}{kp1})}{1/T2 - 1/T1}$

I honestly don't know how to continue.
Could you guide me?
Thank you very much

 

[mjc123]

I think you have to assume that you start with NO₂ and decompose it, i.e. that all the NO and O₂ comes from NO₂, and pNO and pO₂ are not independently variable. Then if you know pNO/pNO₂, you can work out pO₂/pNO₂, and knowing the total pressure, you can work out the individual partial pressures.

 

[il postino]

I think you have to assume that you start with NO₂ and decompose it, i.e. that all the NO and O₂ comes from NO₂, and pNO and pO₂ are not independently variable. Then if you know pNO/pNO₂, you can work out pO₂/pNO₂, and knowing the total pressure, you can work out the individual partial pressures.

Excuse me Mjc, I can't understand yet.