At 50°C the thermodynamic equilibrium constant for the reaction
2NO₂(g) $\leftrightarrows$ N₂O₄(g)
Is 1.18. If a reaction vessel is initially charged with 0.5 atm of NO₂ at this temperature. What will be the partial pressure of N₂O₄ be when the reaction reaches equilibrium?

Question

At 50°C the thermodynamic equilibrium constant for the reaction
2NO₂(g)

\leftrightarrows

N₂O₄(g)
Is 1.18. If a reaction vessel is initially charged with 0.5 atm of NO₂ at this temperature. What will be the partial pressure of N₂O₄ be when the reaction reaches equilibrium?

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is:$$K_p = \frac{P_{N_2O_4}}{P_{NO_2}^2}$$where $P_{N_2O_4}$ and $P_{NO_2}$ are the partial pressures of $N_2O_4$ and $NO_2$, respectively.At equilibrium, the partial pressure of $N_2O_4$ will be:$$P_{N_2O_4} = K_p \times P_{NO_2}^2 = 1.18 \times (0.5 \ atm)^2 = 0.147 \ atm$$Therefore, the correct answer is B) 0.147 atm.

At 50°C the Thermodynamic Equilibrium Constant for the Reaction 2NO2(g) ⇆\leftrightarrows ⇆

At 50°C the Thermodynamic Equilibrium Constant for the Reaction
2NO₂(g) $\leftrightarrows$