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# chm3 - 2 is decreasing at 0.23 M/s at what rate is[H 2 O...

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Consider: 2A B - the rate of A decrease is twice the rate of B increase: Rate A = 2 Rate B - therefore Rate B = ½ Rate A , or Rate = –½ Δ[A]/Δt = Δ[B]/Δt Consider 2A 3B - the rate of A decrease is the rate of B increase: Rate A = Rate B - therefore ½ Rate A = Rate B , or Rate = –½ Δ[A]/Δt = Δ[B]/Δt So, basically for any equation like aA bB - Rate = –1/a Δ[A]/Δt = 1/b Δ[B]/Δt Problem 1 Consider the following reaction: 2 H 2 (g) + O 2 (g) 2 H 2 O(g) 1. Express the rate in terms of changes in each reactant and product with time: - convenient to choose [O 2 ] as the reference since its stoich. coeff. = 1 - for every molecule of O 2 that disappears, 2 molecules of H 2 disappears, so the rate of [O 2 ] decrease is ½ the rate of [H 2 ] decrease and also ½ the rate of [H 2 O] increase: Rate = –½ Δ[H 2 ]/Δt = –Δ[O 2 ]/Δt = ½ Δ[H 2 O]/Δt 2. When [O
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Unformatted text preview: 2 ] is decreasing at 0.23 M/s, at what rate is [H 2 O] increasing?- because [O 2 ] is decreasing, the change in its concentration must be negative, so we substitute the negative value into the expression and solve for Δ[H 2 O]/Δt: ½ Δ[H 2 O]/Δt = –Δ[O 2 ]/Δt = = –(– 0.23 M/s) = 0.23 M/s Δ[H 2 O]/Δt = 2(0.23 M/s) = 0.46 M/s Problem 2 Consider the reaction between NO(g) and O 2 (g) to form N 2 O 3 (g). 1. Express the rate in terms of changes in each reactant and product with time:- first, balance the reaction: 4 NO(g) + O 2 (g) → 2 N 2 O 3 (g)- Rate = –1/4 Δ[NO]/Δt = –Δ[O 2 ]/Δt = ½ Δ[N 2 O 3 ]/Δt 2. How fast is [O 2 ] decreasing when [NO] is decreasing at a rate of 1.60x10-4 M/s?- –Δ[O 2 ]/Δt = –1/4 Δ[NO]/Δt = –1/4 (–1.60x10-4 M/s) = 4.00x10-5 M/s...
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