3 Specify that K refers to rate constants

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

# 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

# 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?