rate law, rate doubled..then?

Question

Given the reaction: 2Mg + O2 → 2MgO with the rate law expressed as k[Mg][O2]^2, I have two questions regarding the effect of changing the concentrations of the reactants on the reaction rate:
a) What would happen to the reaction rate if the concentration of Mg is doubled?
b) What would happen to the reaction rate if the concentration of O2 is doubled?  
For both scenarios, please choose one of the following options: - stay the same - double - triple - quadruple.

Anonymous · Accepted Answer

Rate = k[Mg][O2]^2
If 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.

Anonymous · 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

Anonymous · Answer

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

Dr. Roscoe Ondricka

rate law, rate doubled..then?

3 Answers

AI-Powered Insights

Want to answer this question?

Categories

Tags

Related Questions