Answer:
The value of equilibrium constant at 2000 C is
.
Explanation:
To calculate
of the reaction, we use Van't Hoff equation, which is:
![\ln(\frac{K_2}{K_1})=\frac{\Delta H^o}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%5Eo%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= vapor pressure at temperature ![T_1](https://tex.z-dn.net/?f=T_1)
= vapor pressure at temperature
= Enthalpy change of the reaction= ?
R = Gas constant = 8.314 J/mol K
We have :
![N_2+O_2\rightarrow 2NO](https://tex.z-dn.net/?f=N_2%2BO_2%5Crightarrow%202NO)
Enthalpy of formation of nitrogen gas ,![\Delta H_{f,N_2}= 0kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Bf%2CN_2%7D%3D%200kJ%2Fmol)
Enthalpy of formation of oxygen gas ,![\Delta H_{f,O_2}= 0 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Bf%2CO_2%7D%3D%200%20kJ%2Fmol)
Enthalpy of formation of NO gas ,![]\Delta H^o= 90.3 kJ/mol](https://tex.z-dn.net/?f=%5D%5CDelta%20H%5Eo%3D%2090.3%20kJ%2Fmol)
Enthalpy of the reaction = ![\Delta H^o](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo)
![\Delta H^o=2\times \Delta H^o - (1\times \Delta H_{f,N_2}+1\times \Delta H_{f,O_2})](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D2%5Ctimes%20%5CDelta%20H%5Eo%20-%20%281%5Ctimes%20%5CDelta%20H_%7Bf%2CN_2%7D%2B1%5Ctimes%20%5CDelta%20H_%7Bf%2CO_2%7D%29)
![\Delta H^o=2\times 90.3 kJ/mol - 1\times 0 kJ/mol =1\times 0kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D2%5Ctimes%2090.3%20kJ%2Fmol%20-%201%5Ctimes%200%20kJ%2Fmol%20%3D1%5Ctimes%200kJ%2Fmol)
![\Delta H^o=180.6 kJ/mol = 180,600 J/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D180.6%20kJ%2Fmol%20%3D%20180%2C600%20J%2Fmol)
Equilibrium constant at 25°C = ![K_1=1.95\times 10^{-31}](https://tex.z-dn.net/?f=K_1%3D1.95%5Ctimes%2010%5E%7B-31%7D)
Equilibrium constant at 2,000°C = ![K_2=?](https://tex.z-dn.net/?f=K_2%3D%3F)
![T_1=25^oC=298.15 K, T_2= 2000^oC=2273.15 K](https://tex.z-dn.net/?f=T_1%3D25%5EoC%3D298.15%20K%2C%20T_2%3D%202000%5EoC%3D2273.15%20K)
By using Van't Hoff equation, te
can be calculated:
![\ln(\frac{K_2}{1.95\times 10^{-31}})=\frac{180,600J/mol}{8.314 J/mol K}[\frac{1}{298.15 K}-\frac{1}{2273.15 K}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_2%7D%7B1.95%5Ctimes%2010%5E%7B-31%7D%7D%29%3D%5Cfrac%7B180%2C600J%2Fmol%7D%7B8.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B298.15%20K%7D-%5Cfrac%7B1%7D%7B2273.15%20K%7D%5D)
![K_2=6.045\times 10^{-4}](https://tex.z-dn.net/?f=K_2%3D6.045%5Ctimes%2010%5E%7B-4%7D)
The value of equilibrium constant at 2000 C is
.