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.
ScanSolve AI Answer
Given reaction: H_2(g) + Br_2(g) 2HBr(g) with K_p = 7.1 × 10^4 at 700 K. a) For the reaction 2HBr(g) H_2(g) + Br_2(g): Step 1: Identify the relationship between the given reaction and the target reaction. The target reaction is the reverse of the original reaction. Step 2: Calculate the new K_p. When a reaction is reversed, the new equilibrium constant is the reciprocal of the original equilibrium constant. K_p' = (1)/(K_p) K_p' = (1)/(7.1 × 10^4) K_p' = (1)/(71000) K_p' ≈ 1.408 × 10^-5 The value of K_p is 1.408 × 10^-5. b) For the reaction (1)/(2)H_2(g) + (1)/(2)Br_2(g) HBr(g): Step 1: Identify the relationship between the given reaction and the target reaction. The target reaction is the original reaction multiplied by a factor of (1)/(2). Step 2: Calculate the new K_p. When a reaction is multiplied by a factor 'n', the new equilibrium constant is the original constant raised to the power 'n'. Here, n = (1)/(2). K_p'' = (K_p)^1/2 = sqrt(K_p) K_p'' = sqrt(7.1 × 10^4) K_p'' = sqrt(71000) K_p'' ≈ 266.458 The value of K_p is 266.458. That's 2 down. 3 left today — send the next one.