For the following reaction at 298 K and 1 atm pressure: 2 NO ( g ) + O 2 ( g ) ⇌ 2 N O 2 ( g ) Given: Δ H f ∘ at 298 K are: NO ( g ) = 90.4 k J / m o l and N O 2 ( g ) = 33.8 k J / m o l . Δ S

Question

For the following reaction at 298 K and 1 atm pressure:

$2 NO (g) + O_{2} (g) ⇌ 2 N O_{2} (g)$

Given: $Δ H_{f}^{\circ}$ at 298 K are: $NO (g) = 90.4 k J / m o l$ and $N O_{2} (g) = 33.8 k J / m o l$ .

$Δ S_{r}^{\circ} e a c t i o n$ at 298 K is $- 70.0 J / K$ .

$R = 8.314 J / K / m o l$ .

$2.303 \times 8.314 \times 298 = 5771.25 J / m o l$ .

Then, $lo g K_{p}$ for the reaction is:

TestHub · Accepted Answer

− 2.303 RT lo g K P ​ = Δ r ​ H 0 − T Δ r ​ S 0

Chemistry - Thermodynamics - II Question with Solution | TestHub