The reaction 2NO(g) + O₂(g)$\rightarrow$ 2NO₂(g) Has $\Delta$H_r^$\omicron$ = -4 kJ.mol^-¹. A possible mechanism for this reaction is 2NO(g) Rapid equilibrium,K $f$ N₂O₂(g) N₂O₂(g)+O₂(g)$\rightarrow$2NO₂(g) Slow,K If the activation energy for the reverse reaction is 225 kJ.mol^-¹. What is the activation energy for the forward reaction? A) Because an intermediate is involved, the forward activation energy cannot be calculated with the given information. B) 111 kJ.mol^-¹. C) 339 kJ.mol^-¹. D) 114 kJ.mol^-¹. E) 55.5 kJ.mol^-¹.

Question

Quizplus · Accepted Answer

The activation energy for the forward reaction can be calculated using the given $\Delta H$ for the reaction and the activation energy for the reverse reaction. The activation energy for the forward reaction ($E_a^{\text{forward}}$) can be found by subtracting the reaction enthalpy ($\Delta H$) from the activation energy of the reverse reaction ($E_a^{\text{reverse}}$). Given $\Delta H = -114 \, \text{kJ/mol}$ and $E_a^{\text{reverse}} = 225 \, \text{kJ/mol}$, $E_a^{\text{forward}} = 225 \, \text{kJ/mol} - (-114 \, \text{kJ/mol}) = 111 \, \text{kJ/mol}$.

The Reaction 2NO(g) + O2(g) →\rightarrow→ 2NO2(g) Has Δ\DeltaΔ Hr ο\omicronο = -4 KJ

The Reaction 2NO(g) + O₂(g) $\rightarrow$ 2NO₂(g) Has $\Delta$ H_r^$\omicron$ = -4 KJ