This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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To calculate the equilibrium constant for the target reaction, we need to manipulate the given reactions. Target reaction: O_2(g) + O(g) O_3(g) Given information: 1. NO_2(g) NO(g) + O(g) with K_c1 = 6.8 × 10^-49 2. O_3(g) + NO(g) NO_2(g) + O_2(g) with K_c2 = 5.8 × 10^-34 Step 1: Manipulate the given reactions to obtain the target reaction. We need O_2(g) and O(g) on the reactant side, and O_3(g) on the product side. Reverse reaction (1) to get O(g) on the reactant side: NO(g) + O(g) NO_2(g) The new equilibrium constant for this reversed reaction is K_c1' = (1)/(K_c1). Reverse reaction (2) to get O_2(g) on the reactant side and O_3(g) on the product side: NO_2(g) + O_2(g) O_3(g) + NO(g) The new equilibrium constant for this reversed reaction is K_c2' = (1)/(K_c2). Step 2: Add the manipulated reactions. Add the reversed reaction (1) and the reversed reaction (2): (NO(g) + O(g) NO_2(g)) + (NO_2(g) + O_2(g) O_3(g) + NO(g)) ---------------------------------------------------- NO(g) + O(g) + NO_2(g) + O_2(g) NO_2(g) + O_3(g) + NO(g) Step 3: Cancel common species on both sides. Cancel NO(g) and NO_2(g) from both sides of the equation: O(g) + O_2(g) O_3(g) This is the target reaction. Step 4: Calculate the overall equilibrium constant. When reactions are added, their equilibrium constants are multiplied. K_c = K_c1' × K_c2' K_c = (1)/(K_c1) × (1)/(K_c2) K_c = (1)/((6.8 × 10^-49)) × (5.8 × 10^-34) K_c = (1)/((6.8 × 5.8) × 10^-49-34) K_c = (1)/(39.44 × 10^-83) K_c = (1)/(3.944 × 10^-82) K_c ≈ 0.25355 × 10^82 K_c ≈ 2.5355 × 10^81 Rounding to two significant figures: K_c ≈ 2.5 × 10^81 The equilibrium constant for the reaction is 2.5 × 10^81. 3 done, 2 left today. You're making progress.