[earthnation112]

I have a question regarding rearranging, to give some context, I was just going through a question about theoretical rate laws and trying to make it the same as the experimental determined rate law given by using steady state approximations for the intermediates and then using those expression and substituting them into the theoretical rate law. That was all background information but the question is I don't see how the two equations below mean the same thing:

[tex]J= \frac{-d[H_{2}O_{2}]}{dt} =  (\frac{k_{1}k_{2}}{k_{-1}})[H_{2}O_{2}][H^{+}][X^{-}]+k_{2}[H_{2}O_{2}](\frac{k_{1}}{k_{-1}}).[H^{+}][X^{-}][/tex]

[tex]=(\frac{2k_{1}k_{2}}{k_{-1}})[H_{2}O_{2}][H^{+}][X^{-}]= k_{R}[/tex]

I was told that the expression below are identical:

[tex](\frac{k_{1}k_{2}}{k_{-1}})[H_{2}O_{2}][H^{+}][X^{-}][/tex] and [tex]k_{2}[H_{2}O_{2}](\frac{k_{1}}{k_{-1}}).[H^{+}][X^{-}][/tex]

If there are identical wouldn't it be everything multiplied by two and not just 2k₁k₂?
Would be grateful for any help, if I've missed any crucial information out needed for the question do point it out, thanks.

[Hunter2]

It is not readable. Please format your text again.

[Borek]

It is perfectly readable and nicely formatted. Hunter, check if you have Java script enabled, otherwise you want be able to read the LaTeX equations.

the question is I don't see how the two equations below mean the same thing:

[tex]J= \frac{-d[H_{2}O_{2}]}{dt} =  (\frac{k_{1}k_{2}}{k_{-1}})[H_{2}O_{2}][H^{+}][X^{-}]+k_{2}[H_{2}O_{2}](\frac{k_{1}}{k_{-1}}).[H^{+}][X^{-}][/tex]

[tex]=(\frac{2k_{1}k_{2}}{k_{-1}})[H_{2}O_{2}][H^{+}][X^{-}]= k_{R}[/tex]

Either I am missing something, or it is painfully trivial. Basically it is a+a=2a.

I was told that the expression below are identical:

[tex](\frac{k_{1}k_{2}}{k_{-1}})[H_{2}O_{2}][H^{+}][X^{-}][/tex] and [tex]k_{2}[H_{2}O_{2}](\frac{k_{1}}{k_{-1}}).[H^{+}][X^{-}][/tex]

Sure they are, multiplication is commutative, you can change the order of operands any way you like.

If there are identical wouldn't it be everything multiplied by two and not just 2k₁k₂?

Again, you are either meaning something quite deep, or you have no idea about basics. What do you mean by "everything multiplied, not just 2k₁k₂"? Multiplication is associative, 2×(a×b)=2×a×b, these are equivalent things, so it doesn't matter where the "2" is.

[earthnation112]

It is perfectly readable and nicely formatted. Hunter, check if you have Java script enabled, otherwise you want be able to read the LaTeX equations.

the question is I don't see how the two equations below mean the same thing:

[tex]J= \frac{-d[H_{2}O_{2}]}{dt} =  (\frac{k_{1}k_{2}}{k_{-1}})[H_{2}O_{2}][H^{+}][X^{-}]+k_{2}[H_{2}O_{2}](\frac{k_{1}}{k_{-1}}).[H^{+}][X^{-}][/tex]

[tex]=(\frac{2k_{1}k_{2}}{k_{-1}})[H_{2}O_{2}][H^{+}][X^{-}]= k_{R}[/tex]

Either I am missing something, or it is painfully trivial. Basically it is a+a=2a.

I was told that the expression below are identical:

[tex](\frac{k_{1}k_{2}}{k_{-1}})[H_{2}O_{2}][H^{+}][X^{-}][/tex] and [tex]k_{2}[H_{2}O_{2}](\frac{k_{1}}{k_{-1}}).[H^{+}][X^{-}][/tex]

Sure they are, multiplication is commutative, you can change the order of operands any way you like.

If there are identical wouldn't it be everything multiplied by two and not just 2k₁k₂?

Again, you are either meaning something quite deep, or you have no idea about basics. What do you mean by "everything multiplied, not just 2k₁k₂"? Multiplication is associative, 2×(a×b)=2×a×b, these are equivalent things, so it doesn't matter where the "2" is.

Yeah just went through it and realised how simple it actually was, the confusion arose because I thought the 2 had to be in front of the everything for it to be multiplied by 2 which isn't the case of course, just tried it with a string of number on my calculator and it was obvious why this is the case, I think all the k terms and the concentrations threw me off and I forgot basic maths, again really appreciate the *delete me*