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I have a little bit of confusion about these two, and how they relate to each other, if someone could help me out, it would be great.
For a given reaction "A+B-->AB", the rate of that reaction is given by "rate=k[A]^x*^y" and the exponents on the concentrations of that reaction (ie, the order) must be determined experimentally. There's no correlation between the coefficients of the reaction equation and the order of the rate law.
But then there's the reaction equilibrium. My chem text book explains how you get the Keq by giving a hypothetical reversible reaction "A2(g)+B2(g)<-->2AB" The rate of the forward reaction had to equal the rate of the reverse reaction for the reaction to be in equilibrium, so "rate(f)=rate(r)". Rate of the forward reaction and reverse reaction are each given by their rate laws, so "rate(f)=k(f)*[A2]*[B2]", and "rate(r)=k(r)*[AB]^2". This is the point where flags get raised, because they're not explaining where they got the exponents from. They just told me that I can't extrapolate the exponents of the rate law from the coefficients of the reaction equation, but here they are (arbitrarily?) assuming they are.
Since by definition we have "rate(f)=rate(r)", "k(f)*[A2]*[B2]=k(r)*[AB]^2", and then through a little multiplication and division, we get "k(f)/k(r)=[AB]^2/([A2]*[B2])". I'm then told "Keq=k(f)/k(r)", so we get "Keq=[AB]^2/([A2]*[B2])".
That's reasonable enough, for this one specific equation. But then I'm told that for any reversible reaction "w*W+x*X<-->y*Y+z*Z",
"K=([Y]^y*[Z]^z)/([W]^w*[X]^x)"
When I took general chemistry, I just sort of accepted that, but as I'm reviewing for the MCATs and trying to follow the logic, that makes no sense. The expression for the reaction equilibrium comes from the rate law. The exponents on the rate law cannot be determined from the reaction equation. But apparently the exponents in the expression for Keq can be taken from the coefficients of chemical equation. But those exponents should be the same as the exponents of the rate law...so now I'm confused.
For a given reaction "A+B-->AB", the rate of that reaction is given by "rate=k[A]^x*^y" and the exponents on the concentrations of that reaction (ie, the order) must be determined experimentally. There's no correlation between the coefficients of the reaction equation and the order of the rate law.
But then there's the reaction equilibrium. My chem text book explains how you get the Keq by giving a hypothetical reversible reaction "A2(g)+B2(g)<-->2AB" The rate of the forward reaction had to equal the rate of the reverse reaction for the reaction to be in equilibrium, so "rate(f)=rate(r)". Rate of the forward reaction and reverse reaction are each given by their rate laws, so "rate(f)=k(f)*[A2]*[B2]", and "rate(r)=k(r)*[AB]^2". This is the point where flags get raised, because they're not explaining where they got the exponents from. They just told me that I can't extrapolate the exponents of the rate law from the coefficients of the reaction equation, but here they are (arbitrarily?) assuming they are.
Since by definition we have "rate(f)=rate(r)", "k(f)*[A2]*[B2]=k(r)*[AB]^2", and then through a little multiplication and division, we get "k(f)/k(r)=[AB]^2/([A2]*[B2])". I'm then told "Keq=k(f)/k(r)", so we get "Keq=[AB]^2/([A2]*[B2])".
That's reasonable enough, for this one specific equation. But then I'm told that for any reversible reaction "w*W+x*X<-->y*Y+z*Z",
"K=([Y]^y*[Z]^z)/([W]^w*[X]^x)"
When I took general chemistry, I just sort of accepted that, but as I'm reviewing for the MCATs and trying to follow the logic, that makes no sense. The expression for the reaction equilibrium comes from the rate law. The exponents on the rate law cannot be determined from the reaction equation. But apparently the exponents in the expression for Keq can be taken from the coefficients of chemical equation. But those exponents should be the same as the exponents of the rate law...so now I'm confused.
