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3 Specify that K refers to rate constants
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I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that at equilibriumthe rate constant K is related to the forward and reverse reaction constants: $$K=\frac{K_\mathrm{f}}{K_\mathrm{r}}.$$

In answering the question it is given that: $$K_\mathrm{r}=Ae^{\frac{-(E_\mathrm{a}-\Delta G)}{RT}}$$

And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to $E_\mathrm{a}$ of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?

I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that at equilibrium $$K=\frac{K_\mathrm{f}}{K_\mathrm{r}}.$$

In answering the question it is given that: $$K_\mathrm{r}=Ae^{\frac{-(E_\mathrm{a}-\Delta G)}{RT}}$$

And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to $E_\mathrm{a}$ of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?

I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that the rate constant K is related to the forward and reverse reaction constants: $$K=\frac{K_\mathrm{f}}{K_\mathrm{r}}.$$

In answering the question it is given that: $$K_\mathrm{r}=Ae^{\frac{-(E_\mathrm{a}-\Delta G)}{RT}}$$

And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to $E_\mathrm{a}$ of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?

2 better readable maths
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Forward How to prove that the forward and Reverse Reactionsreverse reactions have the same rate at equilibrium?

I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that at equilibrium $K=\frac{K_{f}}{K_{r}}$ $$K=\frac{K_\mathrm{f}}{K_\mathrm{r}}.$$

In answering the question it is given that:

$K_{r}=Ae^{\frac{-(E_{a}-\Delta G)}{RT}}$ $$K_\mathrm{r}=Ae^{\frac{-(E_\mathrm{a}-\Delta G)}{RT}}$$

And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to Ea$E_\mathrm{a}$ of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?

Forward and Reverse Reactions

I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that at equilibrium $K=\frac{K_{f}}{K_{r}}$

In answering the question it is given that:

$K_{r}=Ae^{\frac{-(E_{a}-\Delta G)}{RT}}$

And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to Ea of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?

How to prove that the forward and reverse reactions have the same rate at equilibrium?

I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that at equilibrium $$K=\frac{K_\mathrm{f}}{K_\mathrm{r}}.$$

In answering the question it is given that: $$K_\mathrm{r}=Ae^{\frac{-(E_\mathrm{a}-\Delta G)}{RT}}$$

And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to $E_\mathrm{a}$ of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?

1
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Forward and Reverse Reactions

I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that at equilibrium $K=\frac{K_{f}}{K_{r}}$

In answering the question it is given that:

$K_{r}=Ae^{\frac{-(E_{a}-\Delta G)}{RT}}$

And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to Ea of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?