rate law, rate doubled..then?

Question

2Mg + O2 => 2MgO k[Mg][O2]^2a)What would happen to the rate if Mg were doubled?b)What would happen to the rate if O2 were doubled?You can choose ; -stay the same. -double. -triple. -quadruple.

Maximus VonRueden · Accepted Answer

Rate = k[Mg][O2]^2If you double [Mg] then the rate will also double. This is called a first-order relationship and occurs when the reactant concentration in the rate equation has no exponent (other than 1).If you double [O2[, the rate will quadruple (4x) since the rate law is second-order with respect to O2. If I tripled [O2], the new rate would be 3^2 = 9x.

Maximus VonRueden · Answer

Tripling the ClO2 concentration from 0.020 to 0.060 causes the rate to increase by 0.0248/0.00276 = 9.0 times, while holding [OH-] constant. R is proportional to [ClO2]^n 9R is propotional to [3ClO2]^n, so n = 2 because 3^2 = 9. The rate is second order with respect to [ClO2] Tripling OH- from 0.030 to 0.090 causes the rate to increase by 0.00828/0.00276 = 3, while holding [ClO2] constant. R is proportional to [OH-]^m 3R is proportional to [3OH-]^m, so m = 1 because 3^1 = 3. The rate is first order with respect to [OH-] . The rate law is R = k[ClO2]^2[OH-]^1 Using trial 1 data: k = R / {[ClO2]^2[OH-]} = 0.0248 / {(0.060)^2(0.030)^1} = 230 1/M^2 R = 230 1/s M^2 x (0.100)^2 x (0.030)^1 = 0.069 M/s

Maximus VonRueden · Answer

hard aspect. browse onto bing and yahoo. just that might help!

Oran Armstrong

rate law, rate doubled..then?

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