Consider the following reaction: 4 HBr(g) + O2(g) 2 H2O(g) + 2 Br2(g) (a) The ra

Question

Consider the following reaction: 4 HBr(g) + O2(g) 2 H2O(g) + 2 Br2(g) (a) The rate law for this reaction is first order in HBr(g) and first order in O2(g). What is the rate law for this reaction? Rate = k [HBr(g)] [O2(g)] Rate = k [HBr(g)]2 [O2(g)] Rate = k [HBr(g)] [O2(g)]2 Rate = k [HBr(g)]2 [O2(g)]2 Rate = k [HBr(g)] [O2(g)]3 Rate = k [HBr(g)]4 [O2(g)] (b) If the rate constant for this reaction at a certain temperature is 7750, what is the reaction rate when [HBr(g)] = 0.00344 M and [O2(g)] = 0.00687 M? Rate = M/s. (c) What is the reaction rate when the concentration of HBr(g) is doubled, to 0.00688 M while the concentration of O2(g) is 0.00687 M? Rate = M/s

Explanation / Answer

let suppose ..... the genral assumption... Doubling the concentration of B (trial 2 vs. trial 1) doubles the rate. R = k [B]^n 2R = k [2B]^n n must be 1 (lst order) Doubling the concentration of A (trial 3 vs. trial 1) quadruples the rate. R = k [A]^m 4R = k [2A]^m m must be 2 (2nd order) R = k [A]^2 [B]^1

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