15.19
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itzeelpadillaDis1A
- Posts: 52
- Joined: Fri Sep 29, 2017 7:03 am
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Alvin Tran 2E
- Posts: 39
- Joined: Fri Sep 29, 2017 7:06 am
Re: 15.19
To find the order of [B] you want to compare reaction 1 and 3 because the concentrations of A and C stay the same.
So we have^{N}(3.02)^{M}(1.25)^{L}}{k(1.25)^{N}(1.25)^{M}(1.25)^{L}})
5.84 = 2.416M
Solve for M and you get 2.
So B is second order.
So we have
5.84 = 2.416M
Solve for M and you get 2.
So B is second order.
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Michelle Lee 2E
- Posts: 64
- Joined: Thu Jul 27, 2017 3:01 am
Re: 15.19
Usually if the numbers aren't so "strange", we would be looking at the exponent to define what order the reactant is.
Say we are talking about doubling the concentration of one reactant while keeping the other reactants constant. If we double the reactant and the rate for that experiment also doubles, 2^n=2, n would equal 1.
Moreover, if we double the concentration of the reactant and the rate for that experiment quadruples, 2^n=4, n would be 2 and the reactant is second order.
We do this for all reactants to see if they show up in the rate law for that chemical reaction.
Say we are talking about doubling the concentration of one reactant while keeping the other reactants constant. If we double the reactant and the rate for that experiment also doubles, 2^n=2, n would equal 1.
Moreover, if we double the concentration of the reactant and the rate for that experiment quadruples, 2^n=4, n would be 2 and the reactant is second order.
We do this for all reactants to see if they show up in the rate law for that chemical reaction.
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