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For N₂O₄(g) ⇌ 2NO₂(g), Kc = 4.61 × 10⁻³ at 298 K…

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For $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$ , $K_c = 4.61 \times 10^{-3}$ at 298 K. For $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ , $K_c = 0.061$ at 500 K. Find $RT$ at 500 K and $K_p$ for both reactions. Use $R = 0.0831$ L bar K $^{-1}$ mol $^{-1}$ .

Need a reminder? The key ideaShow

For $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ : $K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$ . $K_p = K_c(RT)^{\Delta n}$ with $R = 0.0831$ L bar K $^{-1}$ mol $^{-1}$ . Pure solids and liquids are left out. $Q_c < K_c$ : forward; $Q_c > K_c$ : backward. $\Delta G^\circ = -2.303\,RT\log K$ .

The worked solution

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← Write Kc for each reaction, and find Δ n for each: (a) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g); (b)…0.200 mol of PCl₅ is put in a 1.00 L flask and heated… →

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For N₂O₄(g) ⇌ 2NO₂(g), Kc = 4.61 × 10⁻³ at 298 K…

The worked solution

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